2033. Minimum Operations to Make a Uni-Value Grid
Problem Description
The problem presents us with a two-dimensional grid of integers and a single integer x
. The operation that can be executed involves adding or subtracting x
to any element in the grid. A uni-value grid
is defined as a grid where all elements are equal. The goal is to determine the minimum number of such operations required to turn the input grid into a uni-value grid. If this objective is unattainable, the function should return -1
.
The critical constraint to keep in mind is that all elements in the final uni-value grid must be equal, which implies that the difference between any two given elements in the original grid must be a multiple of x
. Otherwise, it would be impossible to reach a common value through the given operation.
Intuition
To arrive at the minimum number of operations needed to achieve a uni-value grid, we need to find a target value that all elements in the grid will be equal to after the operations. Since adding or subtracting x
continues to maintain the same remainder when any element is divided by x
, all elements must share the same remainder mod
when divided by x
; otherwise, there's no legal operation that will create a uni-value grid.
Given that all elements can be modified to share the same remainder when divided by x
, the next step is to decide on which value to set all elements to. A natural choice is the median of all values, as it minimizes the absolute deviation sum for a set of numbers. In other words, using the median value as a target implies the least total difference between each number and the median, hence requiring the fewest operations.
The solution approach follows these steps:
- Flatten the grid into a list
nums
, to work with a single-dimensional array of all the values. - Check if all elements have the same remainder when divided by
x
. If not, return-1
. - Sort
nums
and find the median value. - Sum the absolute division (//) of differences between each element in
nums
and the median byx
. This represents the minimum number of operations to adjust each element to make the entire grid uni-value.
This solution is efficient as it minimally processes the elements by using their inherent properties (specifically, the remainder upon division) and statistical measures (median) to determine feasibility and achieve optimality for the problem.
Solution Approach
The solution approach for converting the grid to a uni-value grid with a minimum number of operations follows a simple but powerful statistical concept. Here is the breakdown of the implementation steps, referencing the provided Python code:
-
Flatten the Grid: The first step involves transforming the two-dimensional grid into a one-dimensional array (list). This is achieved by defining an empty list
nums
and iterating over each row and within that, each valuev
, to append it tonums
. -
Check Modulo Condition: Before proceeding, we need to guarantee that all elements can be adjusted to the same value using the given operation. For this purpose, we store the remainder of the first grid element (e.g.,
grid[0][0] % x
) inmod
and then check ifv % x == mod
for all elements in the grid. If any element doesn't satisfy this, it's impossible to reach a uni-value grid, and the function immediately returns-1
. -
Sort and Find Median: Once we have a flat list
nums
of all values, we sort it to arrange the numbers in ascending order. Finding the median is straightforward from this sorted list; it's the middle value, which minimizes the total number of operations needed. If the list's length is even, either of the two middle values will work due to the way the median minimizes the sum of absolute deviations. The median is found usingnums[len(nums) >> 1]
, which is an efficient way to calculate the median index (the same aslen(nums) // 2
). -
Calculate Total Operations: Finally, we calculate the total number of operations needed by summing the integer division of absolute differences between each element and the median, divided by
x
(i.e.,sum(abs(v - mid) // x for v in nums)
). This represents how many times we'd need to add or subtractx
to/from each element to convert it into the median value. -
Return Result: The result of the sum is the minimum number of operations required to make the grid uni-value, which is what the function returns.
In terms of algorithms and patterns, the problem mainly relies on the properties of numbers (divisibility and modulo operation), and the use of sorting and median as a means to minimize absolute differences. No complex data structures are used beyond the one-dimensional list for sorting and iteration, and the algorithm overall exhibits a time complexity effectively determined by the sorting step, which is typically O(N log N) where N is the number of elements in the grid.
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Start EvaluatorExample Walkthrough
Let's consider a two-dimensional grid and an integer x
with the following values:
Grid: [[2, 4], [6, 8]] x: 2
We want to convert the given grid into a uni-value grid using the minimum number of operations where we can add or subtract x
to any element.
Following the solution approach:
-
Flatten the Grid: We transform our two-dimensional grid into a one-dimensional list
nums
. After flattening,nums
will be[2, 4, 6, 8]
. -
Check Modulo Condition: We check if all elements have the same remainder when divided by
x
. For our grid, all elements (2 % 2
,4 % 2
,6 % 2
,8 % 2
) have a remainder of0
when divided by2
. Since they all have the same remainder, we can proceed. -
Sort and Find Median: We sort the list
nums
to get[2, 4, 6, 8]
. The length of the list is 4 (even), so the median could be either of the two middle values, 4 or 6. Here we'll choose 4 as the median (nums[4 // 2]
). -
Calculate Total Operations: We calculate the number of operations needed to transform each element into the median. For every element
v
innums
, we calculateabs(v - median) // x
and sum these values up:- For
2
:abs(2 - 4) // 2
equals1
operation. - For
4
:abs(4 - 4) // 2
equals0
operations. - For
6
:abs(6 - 4) // 2
equals1
operation. - For
8
:abs(8 - 4) // 2
equals2
operations.
Summing these up gives us
1 + 0 + 1 + 2 = 4
operations. - For
-
Return Result: The minimum number of operations required to make the grid uni-value is
4
. This is the final result we return.
Through these steps, we've used a statistical approach (median) alongside modulo operations to efficiently determine the required operations to reach a uni-value grid, ensuring that the chosen operations are valid and minimal.
Solution Implementation
1class Solution:
2 def minOperations(self, grid, x):
3 """
4 Determine the minimum number of operations required to make all elements in the grid equal,
5 where in one operation you can add or subtract `x` to any element in the grid.
6
7 :param grid: List[List[int]]
8 :param x: int
9 :return: int, the minimum number of operations or -1 if it's not possible
10 """
11
12 # Flatten the grid into a single sorted list to make it easier to handle
13 flattened_grid = []
14
15 # Start by storing the remainder of the first element (for modulo comparison)
16 required_modulo = grid[0][0] % x
17
18 # Iterate through the grid
19 for row in grid:
20 for value in row:
21
22 # If the current value cannot be achieved by any number of operations,
23 # we return -1 because the whole grid cannot be equalized
24 if value % x != required_modulo:
25 return -1
26
27 # Otherwise, store the value in our flattened grid
28 flattened_grid.append(value)
29
30 # Sort all values in our list to easily find the median
31 flattened_grid.sort()
32
33 # Find the median, which will be our target value
34 median = flattened_grid[len(flattened_grid) // 2]
35
36 # Calculate the sum of operations required to make all values equal to the median
37 operations = sum(abs(value - median) // x for value in flattened_grid)
38
39 # Return the total number of operations
40 return operations
41
42# Example usage:
43# solution = Solution()
44# grid = [[1, 5], [2, 3]]
45# x = 1
46# print(solution.minOperations(grid, x)) # Outputs the minimum number of operations required
47
1class Solution {
2 public int minOperations(int[][] grid, int x) {
3 // Get the number of rows
4 int rows = grid.length;
5 // Get the number of columns
6 int cols = grid[0].length;
7 // Initialize an array to store all elements in the grid
8 int[] flattenedGrid = new int[rows * cols];
9 // Modulo of the first element in the grid. All other elements should have the same modulo if a solution is possible
10 int initialMod = grid[0][0] % x;
11
12 // Flatten the 2D grid into a 1D array while checking if operation is not possible
13 for (int i = 0; i < rows; ++i) {
14 for (int j = 0; j < cols; ++j) {
15 // If any element has a different modulo than the first element, return -1 as it's not possible to make them all equal
16 if (grid[i][j] % x != initialMod) {
17 return -1;
18 }
19 // Store elements of the grid in the flattened 1D array
20 flattenedGrid[i * cols + j] = grid[i][j];
21 }
22 }
23
24 // Sort the flattened 1D array
25 Arrays.sort(flattenedGrid);
26 // Find the median element
27 int median = flattenedGrid[flattenedGrid.length / 2];
28 // Initialize operation count to 0
29 int operationCount = 0;
30
31 // Calculate the total number of operations required to make all elements equal to the median
32 for (int value : flattenedGrid) {
33 // Increment the operation count by the number of operations needed for each element.
34 // The distance between the element and the median is divided by x
35 // as that's the number of operations required to make them equal
36 operationCount += Math.abs(value - median) / x;
37 }
38
39 // Return the total operation count
40 return operationCount;
41 }
42}
43
1class Solution {
2public:
3 int minOperations(vector<vector<int>>& grid, int x) {
4 // Calculate the dimensions of the grid
5 int rows = grid.size();
6 int cols = grid[0].size();
7
8 // Determine the modulo of the first element as a reference
9 int referenceModulo = grid[0][0] % x;
10
11 // Flatten the grid into a single vector for ease of manipulation
12 vector<int> flattenedGrid(rows * cols);
13
14 // Process the grid while checking if it's possible to make all elements equal
15 for (int i = 0; i < rows; ++i) {
16 for (int j = 0; j < cols; ++j) {
17 // If an element's modulo isn't equal to the reference, it's not possible to make all elements equal
18 if (grid[i][j] % x != referenceModulo) {
19 return -1;
20 }
21 // Populate the flattened grid
22 flattenedGrid[i * cols + j] = grid[i][j];
23 }
24 }
25
26 // Sort the flattened grid to find the median
27 sort(flattenedGrid.begin(), flattenedGrid.end());
28
29 // The median value is the target value for all elements to make them equal
30 int targetValue = flattenedGrid[(rows * cols) / 2];
31
32 // Calculate the total number of operations needed
33 int totalOps = 0;
34 for (int value : flattenedGrid) {
35 // Add the required number of operations per element
36 totalOps += abs(value - targetValue) / x;
37 }
38
39 // Return the total number of operations
40 return totalOps;
41 }
42};
43
1// Function to find the minimum number of operations needed to make all elements of a grid equal.
2function minOperations(grid: number[][], x: number): number {
3 // Calculate the dimensions of the grid
4 let rows = grid.length;
5 let cols = grid[0].length;
6
7 // Determine the modulo of the first element as a reference
8 let referenceModulo = grid[0][0] % x;
9
10 // Flatten the grid into a single array for ease of manipulation
11 let flattenedGrid: number[] = [];
12
13 // Process the grid while checking if it's possible to make all elements equal
14 for (let i = 0; i < rows; i++) {
15 for (let j = 0; j < cols; j++) {
16 // If an element's modulo isn't equal to the reference, it's not possible to make all elements equal
17 if (grid[i][j] % x !== referenceModulo) {
18 return -1;
19 }
20 // Populate the flattened grid
21 flattenedGrid.push(grid[i][j]);
22 }
23 }
24
25 // Sort the flattened grid to find the median
26 flattenedGrid.sort((a, b) => a - b);
27
28 // The median value is the target value for all elements to make them equal
29 let targetValue = flattenedGrid[Math.floor((rows * cols) / 2)];
30
31 // Calculate the total number of operations needed
32 let totalOps = 0;
33 for (let value of flattenedGrid) {
34 // Add the required number of operations per element
35 totalOps += Math.abs(value - targetValue) / x;
36 }
37
38 // Return the total number of operations
39 return totalOps;
40}
41
Time and Space Complexity
The given Python code is used to determine the minimum number of operations to make all elements in a grid equal if one can only add or subtract a value x
from elements in the grid.
Time Complexity:
The time complexity of the code is analyzed as follows:
- Creating the
nums
list with a nested loop through the grid: This takesO(m*n)
, wherem
andn
are the dimensions of the grid. - Sorting the
nums
list: The sort operation has a time complexity ofO(k*log(k))
, wherek
is the total number of elements in the grid (k = m*n
). - Calculating the median and the number of operations: This involves a single pass through the sorted
nums
list, which takesO(k)
.
Combining these, the dominant term is the sorting step, therefore the overall time complexity is O(k*log(k))
which simplifies to O(m*n*log(m*n))
since k = m*n
.
Space Complexity:
- Storing the
nums
list: Space complexity isO(k)
, which isO(m*n)
wherek
is the total number of elements in the grid. - Variables
mod
andmid
: These are constant space and do not scale with the input, so they contributeO(1)
.
Thus, the overall space complexity is O(m*n)
since the storage for the nums
list dominates the space usage.
Learn more about how to find time and space complexity quickly using problem constraints.
What's the output of running the following function using input [30, 20, 10, 100, 33, 12]
?
1def fun(arr: List[int]) -> List[int]:
2 import heapq
3 heapq.heapify(arr)
4 res = []
5 for i in range(3):
6 res.append(heapq.heappop(arr))
7 return res
8
1public static int[] fun(int[] arr) {
2 int[] res = new int[3];
3 PriorityQueue<Integer> heap = new PriorityQueue<>();
4 for (int i = 0; i < arr.length; i++) {
5 heap.add(arr[i]);
6 }
7 for (int i = 0; i < 3; i++) {
8 res[i] = heap.poll();
9 }
10 return res;
11}
12
1class HeapItem {
2 constructor(item, priority = item) {
3 this.item = item;
4 this.priority = priority;
5 }
6}
7
8class MinHeap {
9 constructor() {
10 this.heap = [];
11 }
12
13 push(node) {
14 // insert the new node at the end of the heap array
15 this.heap.push(node);
16 // find the correct position for the new node
17 this.bubble_up();
18 }
19
20 bubble_up() {
21 let index = this.heap.length - 1;
22
23 while (index > 0) {
24 const element = this.heap[index];
25 const parentIndex = Math.floor((index - 1) / 2);
26 const parent = this.heap[parentIndex];
27
28 if (parent.priority <= element.priority) break;
29 // if the parent is bigger than the child then swap the parent and child
30 this.heap[index] = parent;
31 this.heap[parentIndex] = element;
32 index = parentIndex;
33 }
34 }
35
36 pop() {
37 const min = this.heap[0];
38 this.heap[0] = this.heap[this.size() - 1];
39 this.heap.pop();
40 this.bubble_down();
41 return min;
42 }
43
44 bubble_down() {
45 let index = 0;
46 let min = index;
47 const n = this.heap.length;
48
49 while (index < n) {
50 const left = 2 * index + 1;
51 const right = left + 1;
52
53 if (left < n && this.heap[left].priority < this.heap[min].priority) {
54 min = left;
55 }
56 if (right < n && this.heap[right].priority < this.heap[min].priority) {
57 min = right;
58 }
59 if (min === index) break;
60 [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61 index = min;
62 }
63 }
64
65 peek() {
66 return this.heap[0];
67 }
68
69 size() {
70 return this.heap.length;
71 }
72}
73
74function fun(arr) {
75 const heap = new MinHeap();
76 for (const x of arr) {
77 heap.push(new HeapItem(x));
78 }
79 const res = [];
80 for (let i = 0; i < 3; i++) {
81 res.push(heap.pop().item);
82 }
83 return res;
84}
85
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