Sum of Subarray Minimums in C++

Suppose we have an array of integers A. We have to find the sum of min(B), where B ranges over every (contiguous) subarray of A. Since the answer may be very large, then return the answer in modulo 10^9 + 7. So if the input is like [3,1,2,4], then the output will be 17, because the subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4], so minimums are [3,1,2,4,1,1,2,1,1,1], and the sum is 17.
To solve this, we will follow these steps −
m := 1 x 10^9 + 7
m := 1 x 10^9 + 7
Define two methods, add() will take a, b and returns the (a mod m + b mod m) mod m, mul() will take a, b and returns the (a mod m * b mod m) mod m
Define two methods, add() will take a, b and returns the (a mod m + b mod m) mod m, mul() will take a, b and returns the (a mod m * b mod m) mod m
The main method will take the array A, define a stack st, and set n := size of array A
The main method will take the array A, define a stack st, and set n := size of array A
Define two arrays left of size n and fill with -1, and another is right of size n, fill with n
Define two arrays left of size n and fill with -1, and another is right of size n, fill with n
set ans := 0
set ans := 0
for i in range 0 to n – 1while st is not empty and A[stack top] >= A[i], delete from stif st is not empty, then set left[i] := top of stinsert i into st
for i in range 0 to n – 1
while st is not empty and A[stack top] >= A[i], delete from st
while st is not empty and A[stack top] >= A[i], delete from st
if st is not empty, then set left[i] := top of st
if st is not empty, then set left[i] := top of st
insert i into st
insert i into st
while st is not empty, then delete st
while st is not empty, then delete st
for i in range n – 1 down to 0while st is not empty and A[stack top] >= A[i], delete from stif st is not empty, then set right[i] := top of stinsert i into st
for i in range n – 1 down to 0
while st is not empty and A[stack top] >= A[i], delete from st
while st is not empty and A[stack top] >= A[i], delete from st
if st is not empty, then set right[i] := top of st
if st is not empty, then set right[i] := top of st
insert i into st
insert i into st
for i in range 0 to n – 1leftBound := i – left[i] + 1, rightBound := right[i] – 1 – icontri := 1 + leftBound + rightBound + (leftBound * rightBound)ans := add(ans and mul(contri, A[i]))
for i in range 0 to n – 1
leftBound := i – left[i] + 1, rightBound := right[i] – 1 – i
leftBound := i – left[i] + 1, rightBound := right[i] – 1 – i
contri := 1 + leftBound + rightBound + (leftBound * rightBound)
contri := 1 + leftBound + rightBound + (leftBound * rightBound)
ans := add(ans and mul(contri, A[i]))
ans := add(ans and mul(contri, A[i]))
return ans
return ans
Let us see the following implementation to get better understanding −
 Live Demo
#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
const lli MOD = 1e9 + 7;
class Solution {
public:
   lli add(lli a, lli b){
      return (a % MOD + b % MOD) % MOD;
   }
   lli mul(lli a, lli b){
      return (a % MOD * b % MOD) % MOD;
   }
   int sumSubarrayMins(vector<int>& A) {
      stack <int> st;
      int n = A.size();
      vector <int> left(n, -1);
      vector <int> right(n, n);
      int ans = 0;
      for(int i = 0; i < n; i++){
         while(!st.empty() && A[st.top()] >= A[i]){
         st.pop();
      }
      if(!st.empty())left[i] = st.top();
         st.push(i);
      }
      while(!st.empty())st.pop();
      for(int i = n - 1; i >= 0; i--){
         while(!st.empty() && A[st.top()] > A[i]){
            st.pop();
         }
         if(!st.empty())right[i] = st.top();
            st.push(i);
      }
      for(int i = 0; i < n; i++){
         int leftBound = i - (left[i] + 1);
         int rightBound = (right[i] - 1) - i;
         int contri = 1 + leftBound + rightBound + (leftBound * rightBound);
         ans = add(ans, mul(contri, A[i]));
      }
      return ans;
   }
};
main(){
   vector<int> v = {3,1,2,4};
   Solution ob;
   cout << (ob.sumSubarrayMins(v));
}
[3,1,2,4]
17