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2817. Minimum Absolute Difference Between Elements With Constraint

MediumArrayBinary SearchOrdered Set
Leetcode Link

Problem Description

You have an array of integers nums (0-indexed) and an integer x. Your task is to find the minimum absolute difference between any two elements in the array, with the constraint that these two elements must be at least x indices apart.

More specifically:

  • You need to find two indices i and j where abs(i - j) >= x (the indices are at least x positions apart)
  • Among all such valid pairs, find the one where abs(nums[i] - nums[j]) is minimized
  • Return this minimum absolute difference

For example, if you have an array and x = 3, you can only compare elements that are at least 3 positions apart from each other in the array. You want to find which pair of such elements has the smallest absolute difference between their values.

The solution uses an ordered set (SortedList) to efficiently track elements that are at least x positions away from the current position. As we iterate through the array starting from index x, we:

  1. Add nums[i - x] to the sorted list (this element is exactly x positions behind)
  2. Use binary search to find where nums[i] would fit in the sorted list
  3. Check the closest elements (one just greater than or equal to nums[i], and one just less than nums[i]) to find the minimum absolute difference
  4. Update the answer with the minimum difference found so far
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Intuition

The key insight is that for any element at index i, we need to find the element with minimum absolute difference among all elements that are at least x indices away. A naive approach would check all valid pairs, taking O(n²) time.

To optimize this, we can observe that when we're at index i, all elements from indices 0 to i - x are valid candidates (they're at least x positions behind). As we move forward in the array, more elements become valid candidates.

The critical realization is that to minimize the absolute difference abs(nums[i] - nums[j]), we need to find the element closest in value to nums[i]. In a sorted collection, the closest element to any value must be either:

  1. The smallest element greater than or equal to the target value
  2. The largest element less than the target value

This leads us to maintain a sorted collection of all valid elements (those at least x indices away). When processing element at index i:

  • We add nums[i - x] to our collection (it just became valid, being exactly x positions behind)
  • We use binary search to quickly find where nums[i] would fit in this sorted collection
  • We check the elements immediately before and after this position - these are the closest values to nums[i]
  • The minimum of these differences gives us the best possible match for the current element

By maintaining this sorted structure and using binary search, we reduce the time complexity from O(n²) to O(n log n), making the solution efficient even for large inputs.

Learn more about Binary Search patterns.

Solution Approach

We use an ordered set (implemented as SortedList in Python) to maintain elements that are at least x indices away from the current position. This data structure allows us to:

  • Insert elements in O(log n) time while maintaining sorted order
  • Perform binary search in O(log n) time

The algorithm works as follows:

  1. Initialize: Create an empty SortedList and set the answer to infinity (inf).

  2. Iterate through valid positions: Start from index i = x (the first position where we can have a valid pair) and go through the rest of the array.

  3. Add valid candidates: For each position i, add nums[i - x] to the sorted list. This element is exactly x positions behind the current index, making it a valid candidate for comparison.

  4. Find closest elements using binary search: Use bisect_left(sl, nums[i]) to find the insertion position j where nums[i] would fit in the sorted list. This gives us:

    • sl[j] - the smallest element greater than or equal to nums[i] (if j < len(sl))
    • sl[j - 1] - the largest element less than nums[i] (if j > 0)

  5. Update minimum difference:

    • If there's an element at position j (not out of bounds), calculate sl[j] - nums[i]
    • If there's an element at position j - 1 (when j > 0), calculate nums[i] - sl[j - 1]
    • Update ans with the minimum of these differences

  6. Return result: After processing all elements, return the minimum absolute difference found.

The time complexity is O(n log n) where n is the length of the array, as we perform n - x insertions and binary searches, each taking O(log n) time. The space complexity is O(n) for storing elements in the sorted list.

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Example Walkthrough

Let's walk through the solution with a concrete example to understand how it works.

Example: nums = [1, 5, 3, 19, 18, 25], x = 3

We need to find the minimum absolute difference between elements that are at least 3 indices apart.

Step-by-step execution:

Initialization:

  • sl = SortedList() (empty)
  • ans = inf

Iteration 1: i = 3

  • Current element: nums[3] = 19
  • Add to sorted list: nums[3 - 3] = nums[0] = 1
  • sl = [1]
  • Binary search for 19 in sl: j = bisect_left([1], 19) = 1
  • Check positions:
    • j = 1 is out of bounds (sl has only 1 element)
    • j - 1 = 0: nums[3] - sl[0] = 19 - 1 = 18
  • Update: ans = min(inf, 18) = 18

Iteration 2: i = 4

  • Current element: nums[4] = 18
  • Add to sorted list: nums[4 - 3] = nums[1] = 5
  • sl = [1, 5] (sorted)
  • Binary search for 18 in sl: j = bisect_left([1, 5], 18) = 2
  • Check positions:
    • j = 2 is out of bounds (sl has only 2 elements)
    • j - 1 = 1: nums[4] - sl[1] = 18 - 5 = 13
  • Update: ans = min(18, 13) = 13

Iteration 3: i = 5

  • Current element: nums[5] = 25
  • Add to sorted list: nums[5 - 3] = nums[2] = 3
  • sl = [1, 3, 5] (sorted)
  • Binary search for 25 in sl: j = bisect_left([1, 3, 5], 25) = 3
  • Check positions:
    • j = 3 is out of bounds (sl has only 3 elements)
    • j - 1 = 2: nums[5] - sl[2] = 25 - 5 = 20
  • Update: ans = min(13, 20) = 13

Result: The minimum absolute difference is 13, achieved between nums[1] = 5 and nums[4] = 18 (indices are 3 apart).

This example illustrates how the sorted list grows as we progress through the array, always containing elements that are valid candidates (at least x positions behind). The binary search efficiently finds the closest values to compare with the current element.

Solution Implementation

1from typing import List
2from sortedcontainers import SortedList
3from bisect import bisect_left
4from math import inf
5
6
7class Solution:
8    def minAbsoluteDifference(self, nums: List[int], x: int) -> int:
9        """
10        Find the minimum absolute difference between nums[i] and nums[j]
11        where |i - j| >= x.
12      
13        Args:
14            nums: List of integers
15            x: Minimum index difference required
16      
17        Returns:
18            Minimum absolute difference between valid pairs
19        """
20        # Initialize a sorted list to maintain elements in sorted order
21        sorted_list = SortedList()
22      
23        # Initialize minimum difference to infinity
24        min_diff = inf
25      
26        # Iterate through the array starting from index x
27        # This ensures we can always look back at least x positions
28        for current_idx in range(x, len(nums)):
29            # Add the element that is x positions behind current position
30            # This element becomes a valid candidate for pairing
31            sorted_list.add(nums[current_idx - x])
32          
33            # Find the position where current element would be inserted
34            # in the sorted list (binary search)
35            insert_pos = bisect_left(sorted_list, nums[current_idx])
36          
37            # Check the element at or after the insertion point (if exists)
38            # This gives us the smallest element >= nums[current_idx]
39            if insert_pos < len(sorted_list):
40                min_diff = min(min_diff, sorted_list[insert_pos] - nums[current_idx])
41          
42            # Check the element just before the insertion point (if exists)
43            # This gives us the largest element < nums[current_idx]
44            if insert_pos > 0:
45                min_diff = min(min_diff, nums[current_idx] - sorted_list[insert_pos - 1])
46      
47        return min_diff
48
1class Solution {
2    public int minAbsoluteDifference(List<Integer> nums, int x) {
3        // TreeMap to store elements that are at least x positions away from current index
4        TreeMap<Integer, Integer> treeMap = new TreeMap<>();
5      
6        // Initialize answer with a large value (2^30)
7        int minDifference = 1 << 30;
8      
9        // Iterate through the list starting from index x
10        for (int i = x; i < nums.size(); i++) {
11            // Add the element that is exactly x positions behind current index
12            // merge() increments the count if key exists, otherwise sets count to 1
13            treeMap.merge(nums.get(i - x), 1, Integer::sum);
14          
15            // Find the smallest element >= current element (ceiling)
16            Integer ceilingKey = treeMap.ceilingKey(nums.get(i));
17            if (ceilingKey != null) {
18                // Update minimum difference if ceiling provides smaller difference
19                minDifference = Math.min(minDifference, ceilingKey - nums.get(i));
20            }
21          
22            // Find the largest element <= current element (floor)
23            Integer floorKey = treeMap.floorKey(nums.get(i));
24            if (floorKey != null) {
25                // Update minimum difference if floor provides smaller difference
26                minDifference = Math.min(minDifference, nums.get(i) - floorKey);
27            }
28        }
29      
30        return minDifference;
31    }
32}
33
1class Solution {
2public:
3    int minAbsoluteDifference(vector<int>& nums, int x) {
4        // Initialize the minimum difference with a large value (2^30)
5        int minDiff = 1 << 30;
6      
7        // Use a multiset to maintain sorted elements that are at least x positions apart
8        multiset<int> validElements;
9      
10        // Start from index x, ensuring we can look back at least x positions
11        for (int i = x; i < nums.size(); ++i) {
12            // Add the element that is exactly x positions behind current index
13            // This ensures all elements in the set are at least x positions apart from nums[i]
14            validElements.insert(nums[i - x]);
15          
16            // Find the smallest element >= nums[i] in the set
17            auto lowerBoundIter = validElements.lower_bound(nums[i]);
18          
19            // If such an element exists, update minimum difference
20            if (lowerBoundIter != validElements.end()) {
21                minDiff = min(minDiff, *lowerBoundIter - nums[i]);
22            }
23          
24            // Check the largest element < nums[i] (previous element in the set)
25            if (lowerBoundIter != validElements.begin()) {
26                --lowerBoundIter;
27                minDiff = min(minDiff, nums[i] - *lowerBoundIter);
28            }
29        }
30      
31        return minDiff;
32    }
33};
34
1// Solution to find minimum absolute difference between array elements with index difference >= x
2function minAbsoluteDifference(nums: number[], x: number): number {
3    // Initialize the tree multiset for maintaining sorted elements
4    const multiSet = createTreeMultiSet();
5    const INF = 1 << 30; // Large value representing infinity
6    let minDiff = INF;
7  
8    // Process elements starting from index x
9    for (let i = x; i < nums.length; ++i) {
10        // Add element that is x positions behind current element
11        addToMultiSet(multiSet, nums[i - x]);
12      
13        // Find ceiling (smallest element >= nums[i])
14        const ceilingValue = getCeiling(multiSet, nums[i]);
15        // Find floor (largest element <= nums[i])
16        const floorValue = getFloor(multiSet, nums[i]);
17      
18        // Update minimum difference with ceiling if exists
19        if (ceilingValue !== undefined) {
20            minDiff = Math.min(minDiff, ceilingValue - nums[i]);
21        }
22        // Update minimum difference with floor if exists
23        if (floorValue !== undefined) {
24            minDiff = Math.min(minDiff, nums[i] - floorValue);
25        }
26    }
27  
28    return minDiff;
29}
30
31// Type definition for comparison function
32type Compare<T> = (lhs: T, rhs: T) => number;
33
34// Red-Black Tree Node structure
35interface RBTreeNode<T = number> {
36    data: T;                          // Node value
37    count: number;                    // Count for duplicate values
38    left: RBTreeNode<T> | null;      // Left child
39    right: RBTreeNode<T> | null;     // Right child
40    parent: RBTreeNode<T> | null;    // Parent node
41    color: number;                    // 0 = red, 1 = black
42}
43
44// Red-Black Tree structure
45interface RBTree<T> {
46    root: RBTreeNode<T> | null;      // Root of the tree
47    lt: (l: T, r: T) => boolean;     // Less than comparison function
48}
49
50// Tree-based MultiSet structure
51interface TreeMultiSet<T = number> {
52    _size: number;                    // Total number of elements
53    tree: RBTree<T>;                  // Underlying RB-tree
54    compare: Compare<T>;              // Comparison function
55}
56
57// Create a new RB-tree node
58function createRBTreeNode<T>(data: T): RBTreeNode<T> {
59    return {
60        data: data,
61        left: null,
62        right: null,
63        parent: null,
64        color: 0,  // New nodes are red by default
65        count: 1   // Initialize count to 1
66    };
67}
68
69// Get sibling of a node
70function getSibling<T>(node: RBTreeNode<T>): RBTreeNode<T> | null {
71    if (!node.parent) return null;
72    return isOnLeft(node) ? node.parent.right : node.parent.left;
73}
74
75// Check if node is left child of its parent
76function isOnLeft<T>(node: RBTreeNode<T>): boolean {
77    return node === node.parent!.left;
78}
79
80// Check if node has at least one red child
81function hasRedChild<T>(node: RBTreeNode<T>): boolean {
82    return Boolean(node.left && node.left.color === 0) || 
83           Boolean(node.right && node.right.color === 0);
84}
85
86// Create a new RB-tree
87function createRBTree<T>(compare: Compare<T> = (l: T, r: T) => (l < r ? -1 : l > r ? 1 : 0)): RBTree<T> {
88    return {
89        root: null,
90        lt: (l: T, r: T) => compare(l, r) < 0
91    };
92}
93
94// Left rotation for RB-tree balancing
95function rotateLeft<T>(tree: RBTree<T>, pt: RBTreeNode<T>): void {
96    const rightChild = pt.right!;
97    pt.right = rightChild.left;
98
99    if (pt.right) pt.right.parent = pt;
100    rightChild.parent = pt.parent;
101
102    if (!pt.parent) {
103        tree.root = rightChild;
104    } else if (pt === pt.parent.left) {
105        pt.parent.left = rightChild;
106    } else {
107        pt.parent.right = rightChild;
108    }
109
110    rightChild.left = pt;
111    pt.parent = rightChild;
112}
113
114// Right rotation for RB-tree balancing
115function rotateRight<T>(tree: RBTree<T>, pt: RBTreeNode<T>): void {
116    const leftChild = pt.left!;
117    pt.left = leftChild.right;
118
119    if (pt.left) pt.left.parent = pt;
120    leftChild.parent = pt.parent;
121
122    if (!pt.parent) {
123        tree.root = leftChild;
124    } else if (pt === pt.parent.left) {
125        pt.parent.left = leftChild;
126    } else {
127        pt.parent.right = leftChild;
128    }
129
130    leftChild.right = pt;
131    pt.parent = leftChild;
132}
133
134// Swap colors of two nodes
135function swapColor<T>(node1: RBTreeNode<T>, node2: RBTreeNode<T>): void {
136    const temp = node1.color;
137    node1.color = node2.color;
138    node2.color = temp;
139}
140
141// Swap data of two nodes
142function swapData<T>(node1: RBTreeNode<T>, node2: RBTreeNode<T>): void {
143    const temp = node1.data;
144    node1.data = node2.data;
145    node2.data = temp;
146}
147
148// Fix RB-tree violations after insertion
149function fixAfterInsert<T>(tree: RBTree<T>, pt: RBTreeNode<T>): void {
150    let parent = null;
151    let grandParent = null;
152
153    // Continue fixing while violations exist
154    while (pt !== tree.root && pt.color !== 1 && pt.parent?.color === 0) {
155        parent = pt.parent;
156        grandParent = pt.parent.parent;
157
158        // Case A: Parent is left child of grandparent
159        if (parent === grandParent?.left) {
160            const uncle = grandParent.right;
161
162            // Case 1: Uncle is red - only recoloring needed
163            if (uncle && uncle.color === 0) {
164                grandParent.color = 0;
165                parent.color = 1;
166                uncle.color = 1;
167                pt = grandParent;
168            } else {
169                // Case 2: Node is right child - left rotation needed
170                if (pt === parent.right) {
171                    rotateLeft(tree, parent);
172                    pt = parent;
173                    parent = pt.parent;
174                }
175
176                // Case 3: Node is left child - right rotation needed
177                rotateRight(tree, grandParent);
178                swapColor(parent!, grandParent);
179                pt = parent!;
180            }
181        } else {
182            // Case B: Parent is right child of grandparent
183            const uncle = grandParent!.left;
184
185            // Case 1: Uncle is red - only recoloring needed
186            if (uncle != null && uncle.color === 0) {
187                grandParent!.color = 0;
188                parent.color = 1;
189                uncle.color = 1;
190                pt = grandParent!;
191            } else {
192                // Case 2: Node is left child - right rotation needed
193                if (pt === parent.left) {
194                    rotateRight(tree, parent);
195                    pt = parent;
196                    parent = pt.parent;
197                }
198
199                // Case 3: Node is right child - left rotation needed
200                rotateLeft(tree, grandParent!);
201                swapColor(parent!, grandParent!);
202                pt = parent!;
203            }
204        }
205    }
206    // Root must always be black
207    tree.root!.color = 1;
208}
209
210// Find BST replacement node for deletion
211function findBSTReplacement<T>(x: RBTreeNode<T>): RBTreeNode<T> | null {
212    // When node has 2 children, find inorder successor
213    if (x.left && x.right) return findSuccessor(x.right);
214    // When leaf node
215    if (!x.left && !x.right) return null;
216    // When single child exists
217    return x.left ?? x.right;
218}
219
220// Find inorder successor (leftmost node in right subtree)
221function findSuccessor<T>(x: RBTreeNode<T>): RBTreeNode<T> {
222    let temp = x;
223    while (temp.left) temp = temp.left;
224    return temp;
225}
226
227// Delete a node from RB-tree
228function deleteNode<T>(tree: RBTree<T>, v: RBTreeNode<T>): void {
229    const u = findBSTReplacement(v);
230
231    // Check if both u and v are black
232    const uvBlack = (u === null || u.color === 1) && v.color === 1;
233    const parent = v.parent!;
234
235    if (!u) {
236        // u is null, therefore v is a leaf
237        if (v === tree.root) {
238            tree.root = null;
239        } else {
240            if (uvBlack) {
241                // Both u and v are black - fix double black
242                fixDoubleBlack(tree, v);
243            } else {
244                // Either u or v is red - make sibling red if exists
245                const sibling = getSibling(v);
246                if (sibling) {
247                    sibling.color = 0;
248                }
249            }
250            // Remove v from tree
251            if (isOnLeft(v)) {
252                parent.left = null;
253            } else {
254                parent.right = null;
255            }
256        }
257        return;
258    }
259
260    if (!v.left || !v.right) {
261        // v has exactly 1 child
262        if (v === tree.root) {
263            // v is root - replace with child's data
264            v.data = u.data;
265            v.left = v.right = null;
266        } else {
267            // Detach v and move u up
268            if (isOnLeft(v)) {
269                parent.left = u;
270            } else {
271                parent.right = u;
272            }
273            u.parent = parent;
274            if (uvBlack) {
275                fixDoubleBlack(tree, u);
276            } else {
277                u.color = 1; // Make u black
278            }
279        }
280        return;
281    }
282
283    // v has 2 children - swap with successor and recurse
284    swapData(u, v);
285    deleteNode(tree, u);
286}
287
288// Fix double black violation after deletion
289function fixDoubleBlack<T>(tree: RBTree<T>, x: RBTreeNode<T>): void {
290    if (x === tree.root) return; // Reached root
291
292    const sibling = getSibling(x);
293    const parent = x.parent!;
294  
295    if (!sibling) {
296        // No sibling - push double black up
297        fixDoubleBlack(tree, parent);
298    } else {
299        if (sibling.color === 0) {
300            // Sibling is red
301            parent.color = 0;
302            sibling.color = 1;
303            if (isOnLeft(sibling)) {
304                rotateRight(tree, parent);
305            } else {
306                rotateLeft(tree, parent);
307            }
308            fixDoubleBlack(tree, x);
309        } else {
310            // Sibling is black
311            if (hasRedChild(sibling)) {
312                // Sibling has at least one red child
313                if (sibling.left && sibling.left.color === 0) {
314                    if (isOnLeft(sibling)) {
315                        // Left-left case
316                        sibling.left.color = sibling.color;
317                        sibling.color = parent.color;
318                        rotateRight(tree, parent);
319                    } else {
320                        // Right-left case
321                        sibling.left.color = parent.color;
322                        rotateRight(tree, sibling);
323                        rotateLeft(tree, parent);
324                    }
325                } else {
326                    if (isOnLeft(sibling)) {
327                        // Left-right case
328                        sibling.right!.color = parent.color;
329                        rotateLeft(tree, sibling);
330                        rotateRight(tree, parent);
331                    } else {
332                        // Right-right case
333                        sibling.right!.color = sibling.color;
334                        sibling.color = parent.color;
335                        rotateLeft(tree, parent);
336                    }
337                }
338                parent.color = 1;
339            } else {
340                // Sibling has two black children
341                sibling.color = 0;
342                if (parent.color === 1) {
343                    fixDoubleBlack(tree, parent);
344                } else {
345                    parent.color = 1;
346                }
347            }
348        }
349    }
350}
351
352// Insert a value into RB-tree
353function insertIntoTree<T>(tree: RBTree<T>, data: T): boolean {
354    // Search for insertion position
355    let parent = tree.root;
356    while (parent) {
357        if (tree.lt(data, parent.data)) {
358            if (!parent.left) break;
359            else parent = parent.left;
360        } else if (tree.lt(parent.data, data)) {
361            if (!parent.right) break;
362            else parent = parent.right;
363        } else break;
364    }
365
366    // Insert new node
367    const node = createRBTreeNode(data);
368    if (!parent) {
369        tree.root = node;
370    } else if (tree.lt(node.data, parent.data)) {
371        parent.left = node;
372    } else if (tree.lt(parent.data, node.data)) {
373        parent.right = node;
374    } else {
375        // Duplicate value - increment count
376        parent.count++;
377        return false;
378    }
379    node.parent = parent;
380    fixAfterInsert(tree, node);
381    return true;
382}
383
384// Find a node with given value in RB-tree
385function findInTree<T>(tree: RBTree<T>, data: T): RBTreeNode<T> | null {
386    let p = tree.root;
387    while (p) {
388        if (tree.lt(data, p.data)) {
389            p = p.left;
390        } else if (tree.lt(p.data, data)) {
391            p = p.right;
392        } else {
393            break;
394        }
395    }
396    return p ?? null;
397}
398
399// Create a new TreeMultiSet
400function createTreeMultiSet<T = number>(
401    collection: T[] | Compare<T> = [],
402    compare: Compare<T> = (l: T, r: T) => (l < r ? -1 : l > r ? 1 : 0)
403): TreeMultiSet<T> {
404    // Handle overloaded parameters
405    if (typeof collection === 'function') {
406        compare = collection;
407        collection = [];
408    }
409  
410    const multiSet: TreeMultiSet<T> = {
411        _size: 0,
412        compare: compare,
413        tree: createRBTree(compare)
414    };
415  
416    // Add initial collection elements
417    for (const val of collection) {
418        addToMultiSet(multiSet, val);
419    }
420  
421    return multiSet;
422}
423
424// Add value to multiset
425function addToMultiSet<T>(multiSet: TreeMultiSet<T>, val: T): boolean {
426    const successful = insertIntoTree(multiSet.tree, val);
427    multiSet._size++;
428    return successful;
429}
430
431// Get ceiling value (smallest value >= given value)
432function getCeiling<T>(multiSet: TreeMultiSet<T>, val: T): T | undefined {
433    let p = multiSet.tree.root;
434    let higher = null;
435  
436    // Binary search for ceiling
437    while (p) {
438        if (multiSet.compare(p.data, val) >= 0) {
439            higher = p;
440            p = p.left;
441        } else {
442            p = p.right;
443        }
444    }
445    return higher?.data;
446}
447
448// Get floor value (largest value <= given value)
449function getFloor<T>(multiSet: TreeMultiSet<T>, val: T): T | undefined {
450    let p = multiSet.tree.root;
451    let lower = null;
452  
453    // Binary search for floor
454    while (p) {
455        if (multiSet.compare(val, p.data) >= 0) {
456            lower = p;
457            p = p.right;
458        } else {
459            p = p.left;
460        }
461    }
462    return lower?.data;
463}
464

Time and Space Complexity

Time Complexity: O(n × log n)

The algorithm iterates through the array starting from index x to len(nums), which gives us O(n - x) = O(n) iterations. In each iteration:

  • sl.add(nums[i - x]): Adding an element to a SortedList takes O(log n) time due to the need to maintain sorted order
  • bisect_left(sl, nums[i]): Binary search in the SortedList takes O(log n) time
  • The comparison and update operations (min, array access) take O(1) time

Since we perform O(log n) operations for each of the O(n) iterations, the overall time complexity is O(n × log n).

Space Complexity: O(n)

The SortedList sl can contain at most n - x elements in the worst case (when x = 0, it would contain all n elements). Therefore, the space complexity is O(n) for storing the SortedList. The remaining variables (ans, i, j) use constant space O(1), which doesn't affect the overall space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Error in Index Management

A common mistake is incorrectly calculating which elements are valid candidates for comparison. Developers might accidentally add nums[i - x + 1] or nums[i - x - 1] instead of nums[i - x], or start the loop from x - 1 or x + 1 instead of x.

Why this happens: The constraint |i - j| >= x can be confusing. When at index i, we need elements at indices <= i - x or >= i + x. Since we're iterating forward, we only consider elements behind us.

Solution: Remember that when you're at index i, the element at index i - x is exactly x positions behind, making it the newest valid candidate to add to the sorted list.

2. Forgetting to Handle Edge Cases in Binary Search

When using bisect_left, developers might forget to check bounds before accessing sorted_list[insert_pos] or sorted_list[insert_pos - 1], leading to IndexError.

Incorrect approach:

# This will crash if insert_pos equals len(sorted_list)
min_diff = min(min_diff, sorted_list[insert_pos] - nums[current_idx])

# This will crash if insert_pos equals 0
min_diff = min(min_diff, nums[current_idx] - sorted_list[insert_pos - 1])

Solution: Always check bounds before accessing array elements:

if insert_pos < len(sorted_list):
    min_diff = min(min_diff, sorted_list[insert_pos] - nums[current_idx])

if insert_pos > 0:
    min_diff = min(min_diff, nums[current_idx] - sorted_list[insert_pos - 1])

3. Using Wrong Data Structure

Using a regular list with manual sorting after each insertion (list.append() followed by list.sort()) instead of a proper ordered set implementation.

Why this is problematic: This increases time complexity from O(n log n) to O(n²) because sorting takes O(n) time for each of the n insertions.

Solution: Use SortedList from sortedcontainers or implement a balanced BST/heap-based solution that maintains sorted order efficiently.

4. Incorrect Difference Calculation

Computing abs(sorted_list[insert_pos] - nums[current_idx]) for both cases instead of considering the sign.

Why this matters: Since we know sorted_list[insert_pos] >= nums[current_idx] and sorted_list[insert_pos - 1] < nums[current_idx], we can avoid the abs() function by using:

  • sorted_list[insert_pos] - nums[current_idx] for elements greater than or equal to current
  • nums[current_idx] - sorted_list[insert_pos - 1] for elements less than current

This is both more efficient and clearer in intent.

5. Not Considering All Valid Pairs

Some might try to optimize by only adding elements to the sorted list when i >= 2*x, thinking they need to wait until they can compare with elements both x positions ahead and behind. This misses valid pairs where one element is near the beginning of the array.

Solution: Start adding elements to the sorted list as soon as i = x, ensuring all valid pairs are considered.

Discover Your Strengths and Weaknesses: Take Our 3-Minute Quiz to Tailor Your Study Plan:

What is the best way of checking if an element exists in an unsorted array once in terms of time complexity? Select the best that applies.

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