Sum of Subarray Minimums in C++

C++Server Side ProgrammingProgramming

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

  • 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

  • Define two arrays left of size n and fill with -1, and another is right of size n, fill with n

  • set ans := 0

  • for i in range 0 to n – 1

    • 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

    • insert i into st

  • while st is not empty, then delete st

  • for i in range n – 1 down to 0

    • 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

    • insert i into st

  • for i in range 0 to n – 1

    • leftBound := i – left[i] + 1, rightBound := right[i] – 1 – i

    • contri := 1 + leftBound + rightBound + (leftBound * rightBound)

    • ans := add(ans and mul(contri, A[i]))

  • return ans

Example

Let us see the following implementation to get better understanding −

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;
   }
};

Input

[3,1,2,4]

Output

17
raja
Published on 17-Mar-2020 06:34:33
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