907. Sum of Subarray Minimums

MediumStackArrayDynamic ProgrammingMonotonic Stack

Problem Description

The problem presents a challenge where you're given an array of integers, arr, and you need to calculate the sum of the minimum element for every possible contiguous subarray within arr. Due to potentially large numbers, the result should be returned modulo 10^9 + 7. A contiguous subarray is a sequence of elements from the array that are consecutive with no gaps. For example, in the array [3, 1, 2, 4], [1, 2, 4] is a contiguous subarray, but [3, 2] is not.

Intuition

Solving this problem efficiently requires an understanding that each element of the array will be the minimum in some number of subarrays. So, rather than considering every subarray explicitly, we conceptualize the problem around each element of the array and determine how many subarrays it is the minimum element of.

To arrive at the solution, we must track two things for each element arr[i]:

left[i]: the index of the first smaller element to the left of arr[i]
right[i]: the index of the first element that is less than or equal to arr[i] to the right

With left[i] and right[i] determined, the number of subarrays in which arr[i] is the minimum can be calculated by (i - left[i]) * (right[i] - i).

Notice that we are specifically looking for the first element to the right that is less than or equal to arr[i] to prevent double-counting certain subarrays. If we used a strictly less than condition, subarrays with same minimum values at different positions could be considered twice.

To maintain these left and right indices, we use a monotonic stack, which is a stack that keeps elements in either increasing or decreasing order. The monotonic stack is traversed twice: once from left to right to find each left[i], and once from right to left to find each right[i]. Each element in the stack represents the next greater element for elements not yet encountered or processed.

The product of (i - left[i]) and (right[i] - i) gives us the count of subarrays where arr[i] is the minimum. This count is then multiplied by arr[i] to get the contribution to the sum from arr[i]. Finally, we sum up the contributions from all elements and return it modulo 10^9 + 7 to handle the large number possibility.

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Solution Approach

The implementation leverages two main concepts: the "Monotonic Stack" pattern and the "Prefix Sum" pattern, to efficiently solve the problem without having to evaluate every subarray explicitly. Here's the walk-through of the implemented solution, step-by-step:

Initialize two arrays, left and right, of the same length as arr to n, with -1 and n respectively. These arrays will hold for each element the index of the previous smaller element (left) and the next smaller or equal element (right).
Initialize a stack stk which we'll use to iterate over the array to find the left and right indices. The stack approach efficiently maintains a decreasing order of elements and their indices.
Iterate through the elements of arr from left to right. For each element, while the stack is not empty and the top element of the stack is greater than or equal to the current element, pop elements from the stack. This process is maintaining the stack in a strictly decreasing order.
After elements larger than the current one are popped off stk, if the stack is not empty, set left[i] to the index of the top element of stk, which is the closest previous element smaller than arr[i]. Then, push the current index i onto the stack.
Clear the stack and then iterate through the elements of arr from right to left to similarly identify right[i] for each element. The process mirrors step 3 and 4, but in the reversed direction and with the condition that any equal value element could also terminate the loop, maintaining strict decreasing order up to equal values.
Once both left and right arrays are filled with proper indices, calculate the sum. By iterating over all indices i, find the product of the count of subarrays where arr[i] is the minimum ((i - left[i]) * (right[i] - i)) and arr[i]. This represents the sum of arr[i] for all i in its valid subarrays.
Sum these products for all i. As the final sum might be very large, each addition is taken modulo 10^9 + 7 to prevent integer overflow.

By combining the Monotonic Stack to find bounds for each element and the Prefix Sum pattern to calculate each element's contribution to the total sum, the algorithm achieves an efficient solution that operates in O(n) time complexity, where n is the size of the input array arr.

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Example Walkthrough

Let's consider the array arr = [3, 1, 2, 4] and walk through the solution method to understand how the implemented algorithm works step-by-step:

First, initialize two arrays, left as [-1, -1, -1, -1] and right as [4, 4, 4, 4]. The -1 indicates that we didn't find a previous smaller element yet, and 4 is used because it's the size of arr, indicating we haven't found the next smaller or equal element yet.
Initialize an empty stack stk to keep track of the indices of the elements we browse through.
Iterate through arr from left to right:
- i = 0, arr[i] = 3, stack is empty, so push 0 onto the stack.
- i = 1, arr[i] = 1, stack top element is 3 (> 1), so pop 0. left[1] becomes the previous element, -1. Stack is empty, push 1.
- i = 2, arr[i] = 2, stack top element is 1 (< 2), do nothing. Push 2 to the stack.
- i = 3, arr[i] = 4, stack top element is 2 (< 4), do nothing. Push 3 to the stack.
After this loop, left becomes [-1, -1, 1, 2], and stack stk contains [1, 2, 3].
Clear the stack for the next phase.
Iterate through arr from right to left to fill the right array:
- i = 3, arr[i] = 4, stack is empty, push 3 onto the stack.
- i = 2, arr[i] = 2, stack top element is 4 (> 2), so pop 3. Push 2 onto the stack.
- i = 1, arr[i] = 1, stack top element is 2 (> 1), pop 2, right[1] is 2. Stack is now empty again, push 1.
- i = 0, arr[i] = 3, stack top element is 1 (< 3), so do nothing. Push 0 onto the stack.
After this loop, right becomes [4, 2, 4, 4], and stack stk contains [1, 0].
Now, the sum is calculated. Iterate over the indices i and for each:
- i = 0, (i - left[i]) * (right[i] - i) -> (0 - (-1)) * (4 - 0) -> 1 * 4 = 4 -> 4 * 3 = 12.
- i = 1, (i - left[i]) * (right[i] - i) -> (1 - (-1)) * (2 - 1) -> 2 * 1 = 2 -> 2 * 1 = 2.
- i = 2, (i - left[i]) * (right[i] - i) -> (2 - 1) * (4 - 2) -> 1 * 2 = 2 -> 2 * 2 = 4.
- i = 3, (i - left[i]) * (right[i] - i) -> (3 - 2) * (4 - 3) -> 1 * 1 = 1 -> 1 * 4 = 4.
The total sum is 12 + 2 + 4 + 4 = 22.
The result would be the sum 22 modulo 10^9 + 7, which is simply 22, as it's less than the modulo value.

By following the steps outlined in the solution, we efficiently find the sum without ever calculating each subarray's minimum explicitly. This example demonstrates how both Monotonic Stack and Prefix Sum patterns can be harnessed to solve a seemingly complex array problem with an elegant and efficient algorithm.

Solution Implementation

1# The Solution class contains a method to find the sum of minimums of all subarrays in a given array.
2class Solution:
3    def sumSubarrayMins(self, arr: List[int]) -> int:
4        n = len(arr)  # Get the length of the input array
5        left = [-1] * n  # Store the index of previous less element for each element in the array
6        right = [n] * n  # Store the index of next less element for each element in the array
7        stack = []  # Initialize an empty stack for indices
8
9        # Calculate the previous less element for each element in the array
10        for i, value in enumerate(arr):
11            while stack and arr[stack[-1]] >= value:  # Ensure that the top of the stack is < current element value
12                stack.pop()  # Pop elements from stack while the current element is smaller or equal
13            if stack:
14                left[i] = stack[-1]  # Update left index if stack is not empty
15            stack.append(i)  # Push the current index onto the stack
16
17        stack = []  # Reset the stack for the next loop
18
19        # Calculate the next less element for each element in the array, going backwards
20        for i in range(n - 1, -1, -1):  # Start from end of the array and move backwards
21            while stack and arr[stack[-1]] > arr[i]:  # Similar stack operation but with strict inequality
22                stack.pop()  # Pop elements while current element is smaller
23            if stack:
24                right[i] = stack[-1]  # Update right index if stack is not empty
25            stack.append(i)  # Push current index to the stack
26
27        mod = 10**9 + 7  # Define modulus for the final result
28
29        # Calculate the sum of all minimum subarray values with their respective frequencies
30        # The frequency is the product of the lengths of subarrays to the left and right
31        result = sum((i - left[i]) * (right[i] - i) * value for i, value in enumerate(arr)) % mod
32      
33        return result  # Return the result sum, modulo 10^9 + 7
34

1class Solution {
2    public int sumSubarrayMins(int[] arr) {
3        int length = arr.length;
4        // Arrays to keep track of previous smaller and next smaller elements
5        int[] left = new int[length];
6        int[] right = new int[length];
7      
8        // Initialize left array with -1 indicating the start of array
9        Arrays.fill(left, -1);
10        // Initialize right array with length of array indicating the end of array
11        Arrays.fill(right, length);
12      
13        // Stack to keep track of elements while traversing
14        Deque<Integer> stack = new ArrayDeque<>();
15      
16        // Calculate previous smaller elements for each element in the array
17        for (int i = 0; i < length; ++i) {
18            // Popping all elements which are greater than the current element
19            while (!stack.isEmpty() && arr[stack.peek()] >= arr[i]) {
20                stack.pop();
21            }
22            // The current top of the stack indicates the previous smaller element
23            if (!stack.isEmpty()) {
24                left[i] = stack.peek();
25            }
26            // Push the current index into the stack
27            stack.push(i);
28        }
29      
30        // Clear the stack for next traversal
31        stack.clear();
32      
33        // Calculate next smaller elements for each element in the array in reverse order
34        for (int i = length - 1; i >= 0; --i) {
35            // Popping all elements which are greater than or equal to the current element
36            while (!stack.isEmpty() && arr[stack.peek()] > arr[i]) {
37                stack.pop();
38            }
39            // The current top of the stack indicates the next smaller element
40            if (!stack.isEmpty()) {
41                right[i] = stack.peek();
42            }
43            // Push the current index into the stack
44            stack.push(i);
45        }
46      
47        // The mod value for big integer operations to prevent overflow
48        int mod = (int) 1e9 + 7;
49        // The result variable to keep track of the sum of subarray minimums
50        long answer = 0;
51      
52        // Calculate the contribution of each element as a minimum in its possible subarrays
53        for (int i = 0; i < length; ++i) {
54            // Total count of subarrays where arr[i] is min is (i - left[i]) * (right[i] - i)
55            // Multiply the count by the value itself and apply modulo operation
56            answer += (long) (i - left[i]) * (right[i] - i) % mod * arr[i] % mod;
57            // Ensure the running sum doesn't overflow
58            answer %= mod;
59        }
60      
61        // Cast the long result to int before returning as per method return type
62        return (int) answer;
63    }
64}
65

1using ll = long long;  // Define 'll' as an alias for 'long long' type
2const int MOD = 1e9 + 7;  // The modulo value
3
4class Solution {
5public:
6    // Function to calculate the sum of minimum elements in every subarray
7    int sumSubarrayMins(vector<int>& nums) {
8        int length = nums.size();  // Get the number of elements in the array
9        vector<int> left(length, -1);  // Create a vector to store indices of previous less element
10        vector<int> right(length, length);  // Create a vector to store indices of next less element
11        stack<int> stk;  // Stack to help find previous and next less elements
12      
13        // Finding previous less element for each index
14        for (int i = 0; i < length; ++i) {
15            while (!stk.empty() && nums[stk.top()] >= nums[i]) {
16                stk.pop();  // Pop elements that are greater or equal to current element
17            }
18            if (!stk.empty()) {
19                left[i] = stk.top();  // Store the index of the previous less element
20            }
21            stk.push(i);  // Push the current index onto the stack
22        }
23      
24        // Clear stack for reuse
25        stk = stack<int>();
26      
27        // Finding next less element for each index
28        for (int i = length - 1; i >= 0; --i) {
29            while (!stk.empty() && nums[stk.top()] > nums[i]) {
30                stk.pop();  // Pop elements that are strictly greater than current element
31            }
32            if (!stk.empty()) {
33                right[i] = stk.top();  // Store the index of the next less element
34            }
35            stk.push(i);  // Push the current index onto the stack
36        }
37      
38        ll sum = 0;  // Initialize the sum of minimum elements in all subarrays
39      
40        // Calculating the contribution of each element to the overall sum
41        for (int i = 0; i < length; ++i) {
42            sum += static_cast<ll>(i - left[i]) * (right[i] - i) * nums[i] % MOD;
43            sum %= MOD;  // Apply modulus to keep the sum within range
44        }
45      
46        return sum;  // Return the sum of minimums of all subarrays
47    }
48};
49

1function sumSubarrayMins(arr: number[]): number {
2    const n = arr.length; // Get the length of the array
3
4    // Helper function to get the element at index i or return a sentinel value for boundaries
5    function getElement(i: number): number {
6        if (i === -1 || i === n) return Number.MIN_SAFE_INTEGER; // Using safe min value for bounds
7        return arr[i];
8    }
9
10    let answer = 0; // Initialize accumulator for the answer
11    const MOD = 1e9 + 7; // The modulo value to prevent integer overflow
12    let stack: number[] = []; // Initialize an empty stack to store indices
13
14    // Iterate through all elements including boundaries
15    for (let i = -1; i <= n; i++) {
16        // While there are elements on the stack and the current element is smaller than the last
17        // on the stack, process the stack and update the answer.
18        while (stack.length && getElement(stack[0]) > getElement(i)) {
19            const index = stack.shift()!; // Remove the top element of the stack
20            // Calculate the contribution of the subarrays where arr[index] is the minimum
21            // and add it to the answer
22            answer = (answer + arr[index] * (index - stack[0]) * (i - index)) % MOD;
23        }
24        // Push the current index onto the stack
25        stack.unshift(i);
26    }
27
28    return answer; // Return the final answer
29}
30
31// Note: The stack is being used to maintain a list of indices in non-decreasing order.
32// By ensuring this ordering, we can efficiently find the previous and next smaller elements
33// for every element in the array, which is essential for finding the minimum of each subarray.
34

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Time and Space Complexity

Time Complexity

The time complexity of the code is O(n), which corresponds to the reference answer. Here's a breakdown of the main operations and their complexities:

Initializing the left and right arrays takes O(n) time as it fills them with default values based on the length of the array arr.
The first for-loop to populate the left array processes each element in arr once, resulting in O(n) time complexity. Even though there's a while-loop inside, it won't lead to a complexity higher than O(n) because elements are only added to and removed from the stack once.
The second for-loop to populate the right array also runs in O(n) time for the same reasons as the first loop.
Finally, the sum computation with a list comprehension operates over each index once, yielding a time complexity of O(n).

Combining these operations, which all run sequentially and independently, the total time complexity remains O(n).

Space Complexity

The space complexity of the code is O(n):

Two additional arrays left and right of size n are created, contributing 2n to the space complexity.
A stack is used, which, in the worst case, could also store up to n elements. However, the stack is reused and not stored in memory all at once. Each element is pushed and popped once.
The list comprehension does not create a new list; it merely iterates over the array to compute the sum, so it does not add to the space complexity.

As none of the auxiliary data structures' sizes exceed n, the total space complexity is O(n).

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907. Sum of Subarray Minimums

Problem Description

Intuition

Solution Approach

Example Walkthrough

Solution Implementation

Time and Space Complexity

Time Complexity

Space Complexity

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