Maximum Sum Circular Subarray in C++


Suppose we have a circular array C of integers represented by A, we have to find the maximum possible sum of a non-empty subarray of C. Also, a subarray may only include each element of the fixed buffer A at most once. If the array is like [1,-2,3,-2], then the output will be 3. This is because subarray[3] has maximum sum 3.

To solve this, we will follow these steps −

  • n := size of v

  • create arrays leftSum, leftSumMax, rightSum, rightSumMax all of size n

  • leftSum[0] := v[0], leftSumMax[0] := maximum of 0 and v[0]

  • for i in range 1 to n – 1

    • leftSum[i] := leftSum[i - 1] + v[i]

    • leftSumMax[i] := maximum of leftSum[i] and leftSumMax[i - 1]

  • rightSum[n - 1] := v[n - 1], leftSumMax[n - 1] := maximum of 0 and v[n - 1]

  • for i in range n - 2 down to 0

    • rightSum[i] := rightSum [i + 1] + v[i]

    • rightSumMax[i] := maximum of rightSum[i + 1] and rightSum Max[i]

  • leftAns := leftSum[0] + rightSumMax[1]

  • for i in range 1 to n – 2

    • leftAns := maximum of leftAns, leftSum[i] + rightSumMax[i + 1]

  • rightAns := rightSum[n - 1] + leftSumMax[n - 2]

  • for i in range n - 2 down to 1

    • rightAns := maximum of rightAns, rightSum[i] + leftSumMax[i - 1]

  • curr := v[0], kadane := v[0]

  • for i in range 1 to n – 1

    • curr := max of v[1], curr + v[i]

    • kadane := max of curr and kadane

  • return the max of leftAns, rightAns and kadane

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int maxSubarraySumCircular(vector<int>& v) {
      int n = v.size();
      vector <int> leftSum(n),leftSumMax(n),rightSum(n), rightSumMax(n);
      leftSum[0] = v[0];
      leftSumMax[0] = max((int)0,v[0]);
      for(int i =1;i<n;i++){
         leftSum[i] = leftSum[i-1] + v[i];
         leftSumMax[i] = max(leftSum[i],leftSumMax[i-1]);
      }
      rightSum[n-1] = v[n-1];
      rightSumMax[n-1] = max((int)0,v[n-1]);
      for(int i =n-2;i>=0;i--){
         rightSum[i] = rightSum[i+1]+v[i];
         rightSumMax[i] = max(rightSumMax[i+1],rightSum[i]);
      }
      int leftAns=leftSum[0]+rightSumMax[1];
      for(int i =1;i<n-1;i++){
         leftAns = max(leftAns,leftSum[i]+rightSumMax[i+1]);
      }
      int rightAns = rightSum[n-1]+leftSumMax[n-2];
      for(int i =n-2;i>=1;i--){
         rightAns = max(rightAns,rightSum[i]+leftSumMax[i-1]);
      }
      int curr=v[0];
      int kadane = v[0];
      for(int i =1;i<n;i++){
         curr = max(v[i],curr+v[i]);
         kadane = max(curr,kadane);
      }
      return max(leftAns,max(rightAns,kadane));
   }
};
main(){
   vector<int> v = {1,-2,3,-2};
   Solution ob;
   cout << (ob.maxSubarraySumCircular(v));
}

Input

[1,-2,3,-2]

Output

3

Updated on: 02-May-2020

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