# 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

min([1, 2]) = 1

min([1, 2, 4]) = 1

min([1, 2, 4, 3]) = 1

min() = 2

min([2, 4]) = 2

min([2, 4, 3]) = 2

min() = 4

min([4, 3]) = 3

min() = 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);
}

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]:
currentSum -= s[-1]
s.pop()
if not s:
s.append([index, value, (index + 1) * value])
else:
s.append([index, value, (index - s[-1]) * value])
currentSum += s[-1]
ans += currentSum
return ans % (10 ** 9 + 7)```
```

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