907. Sum of Subarray Minimums LeetCode Solution

In this guide, you will get 907. Sum of Subarray Minimums LeetCode Solution with the best time and space complexity. The solution to Sum of Subarray Minimums problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Sum of Subarray Minimums solution in C++
4. Sum of Subarray Minimums solution in Java
5. Sum of Subarray Minimums solution in Python
6. Additional Resources

Problem Statement of Sum of Subarray Minimums

Given an array of integers arr, find the sum of min(b), where b ranges over every (contiguous) subarray of arr. Since the answer may be large, return the answer modulo 109 + 7.

Example 1:

Input: arr = [3,1,2,4]
Output: 17
Explanation:
Subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4].
Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1.
Sum is 17.

Example 2:

Input: arr = [11,81,94,43,3]
Output: 444

Constraints:

1 <= arr.length <= 3 * 104
1 <= arr[i] <= 3 * 104

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(n)

907. Sum of Subarray Minimums LeetCode Solution in C++

class Solution {
 public:
  int sumSubarrayMins(vector<int>& arr) {
    constexpr int kMod = 1'000'000'007;
    const int n = arr.size();
    long ans = 0;
    // prevMin[i] := index k s.t. arr[k] is the previous minimum in arr[:i]
    vector<int> prevMin(n, -1);
    // nextMin[i] := index k s.t. arr[k] is the next minimum in arr[i + 1:]
    vector<int> nextMin(n, n);
    stack<int> stack;

    for (int i = 0; i < n; ++i) {
      while (!stack.empty() && arr[stack.top()] > arr[i]) {
        const int index = stack.top();
        stack.pop();
        nextMin[index] = i;
      }
      if (!stack.empty())
        prevMin[i] = stack.top();
      stack.push(i);
    }

    for (int i = 0; i < n; ++i) {
      ans += static_cast<long>(arr[i]) * (i - prevMin[i]) * (nextMin[i] - i);
      ans %= kMod;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

907. Sum of Subarray Minimums LeetCode Solution in Java

class Solution {
  public int sumSubarrayMins(int[] arr) {
    final int kMod = 1_000_000_007;
    final int n = arr.length;
    long ans = 0;
    // prevMin[i] := index k s.t. arr[k] is the previous minimum in arr[:i]
    int[] prevMin = new int[n];
    // nextMin[i] := index k s.t. arr[k] is the next minimum in arr[i + 1:]
    int[] nextMin = new int[n];
    Deque<Integer> stack = new ArrayDeque<>();

    Arrays.fill(prevMin, -1);
    Arrays.fill(nextMin, n);

    for (int i = 0; i < arr.length; ++i) {
      while (!stack.isEmpty() && arr[stack.peek()] > arr[i]) {
        final int index = stack.pop();
        nextMin[index] = i;
      }
      if (!stack.isEmpty())
        prevMin[i] = stack.peek();
      stack.push(i);
    }

    for (int i = 0; i < arr.length; ++i) {
      ans += (long) arr[i] * (i - prevMin[i]) * (nextMin[i] - i);
      ans %= kMod;
    }

    return (int) ans;
  }
}
// code provided by PROGIEZ

907. Sum of Subarray Minimums LeetCode Solution in Python

class Solution:
  def sumSubarrayMins(self, arr: list[int]) -> int:
    kMod = 1_000_000_007
    n = len(arr)
    ans = 0
    # prevMin[i] := index k s.t. arr[k] is the previous minimum in arr[:i]
    prevMin = [-1] * n
    # nextMin[i] := index k s.t. arr[k] is the next minimum in arr[i + 1:]
    nextMin = [n] * n
    stack = []

    for i, a in enumerate(arr):
      while stack and arr[stack[-1]] > a:
        index = stack.pop()
        nextMin[index] = i
      if stack:
        prevMin[i] = stack[-1]
      stack.append(i)

    for i, a in enumerate(arr):
      ans += a * (i - prevMin[i]) * (nextMin[i] - i)
      ans %= kMod

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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