**Every Sublist Min Sum- Google Top Interview Questions**

### Problem Statement :

You are given a list of integers nums. Return the sum of min(x) for every sublist x in nums. Mod the result by 10 ** 9 + 7. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [1, 2, 4, 3] Output 20 Explanation We have the following sublists and their mins: min([1]) = 1 min([1, 2]) = 1 min([1, 2, 4]) = 1 min([1, 2, 4, 3]) = 1 min([2]) = 2 min([2, 4]) = 2 min([2, 4, 3]) = 2 min([4]) = 4 min([4, 3]) = 3 min([3]) = 3

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums) {
int n = nums.size();
int pref[n + 1] = {};
stack<int> s; // Initialising prefix array and stack
int ans = 0, j = 0, mod = 1e9 + 7;
for (int i = 0; i < n; i++) {
while (!s.empty() && nums[s.top()] >= nums[i])
s.pop(); // iterating back till we find an index where the minimum of the list if we go
// any further left will change
if (s.empty())
j = -1; // we are calculating j just for calculating the count for which nums[i] will
// appear as minimum in the sublist we have created by going left
else
j = s.top();
s.push(i);
pref[i + 1] = (pref[j + 1] + (i - j) * nums[i]) %
mod; // here maintaining prefix array--Read the implementation part
ans = (ans + pref[i + 1]) % mod;
}
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums) {
int n = nums.length;
int[] next_smaller = new int[n];
int[] prev_smaller = new int[n];
Arrays.fill(next_smaller, n);
Arrays.fill(prev_smaller, -1);
// compute and fill in the next_smaller
Stack<Integer> stack = new Stack();
for (int i = 0; i < n; i++) {
int cur = nums[i];
while (stack.size() > 0 && cur < nums[stack.peek()]) {
next_smaller[stack.pop()] = i;
}
stack.push(i);
}
// compute and fill in the prev_smaller
stack = new Stack();
for (int i = 0; i < n; i++) {
int cur = nums[i];
while (stack.size() > 0 && nums[stack.peek()] > cur) {
stack.pop();
}
if (stack.size() > 0)
prev_smaller[i] = stack.peek();
stack.push(i);
}
// compute the answer
long res = 0;
long mod = (long) 1e9 + 7;
for (int i = 0; i < n; i++) {
int subarray_min = nums[i];
int left_count = i - prev_smaller[i]; // number of greater elements on the left
int right_count =
next_smaller[i] - i; // number of greater than or equal to elements on right
res = res + (subarray_min * (left_count * right_count)) % mod;
res = res % mod;
}
return (int) res;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
ans = 0
s = [] # stack of (index, value, sum)
currentSum = 0
for index, value in enumerate(nums):
while s and value <= s[-1][1]:
currentSum -= s[-1][2]
s.pop()
if not s:
s.append([index, value, (index + 1) * value])
else:
s.append([index, value, (index - s[-1][0]) * value])
currentSum += s[-1][2]
ans += currentSum
return ans % (10 ** 9 + 7)
```

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