Merge pull request #435 from RAUNAK-PANDEY/Raunak

minimum difference of two subset using C++ #415

Author: smv1999

Dynamic Programming/minimum_diffrenece subset.cpp

@@ -0,0 +1,90 @@
+//Link : https://www.interviewbit.com/problems/minimum-difference-subsets/
+/*
+Problem Description
+
+Given an integer array A containing N integers.
+
+You need to divide the array A into two subsets S1 and S2 such that the absolute difference between their sums is minimum.
+
+Find and return this minimum possible absolute difference.
+
+NOTE:
+
+Subsets can contain elements from A in any order (not necessary to be contiguous).
+Each element of A should belong to any one subset S1 or S2, not both.
+It may be possible that one subset remains empty.
+
+
+Problem Constraints
+1 <= N <= 100
+
+1 <= A[i] <= 100
+
+
+
+Input Format
+First and only argument is an integer array A.
+
+
+
+Output Format
+Return an integer denoting the minimum possible difference among the sums of two subsets.
+
+
+
+Example Input
+Input 1:
+
+ A = [1, 6, 11, 5]
+
+
+Example Output
+Output 1:
+
+ 1
+
+
+Example Explanation
+Explanation 1:
+
+ Subset1 = {1, 5, 6}, sum of Subset1 = 12
+ Subset2 = {11}, sum of Subset2 = 11*/
+
+int Solution::solve(vector<int> &A) {
+    int sum = 0,N=A.size();
+    for (int i = 0; i < A.size(); i++)
+        sum += A[i];
+
+    bool dp[N+1][sum+1];
+        
+        for(int i=0;i<N+1;i++)
+        {
+            for(int j=0;j<sum+1;j++)
+            {
+                 
+                if(j==0) dp[i][j]= true;
+                else if(i==0) dp[i][j]= false;
+                 
+                else if(A[i-1] <= j)
+                {
+                    dp[i][j] = dp[i-1][j-A[i-1]] || dp[i-1][j];
+                }
+                else dp[i][j] = dp[i-1][j];
+            }
+        }
+            // Initialize difference of two sums.
+        int diff = INT_MAX;
+        
+        // Find the largest j such that dp[n][j]
+        // is true where j loops from sum/2 t0 0
+        for (int j=sum/2; j>=0; j--)
+        {
+            // Find the
+            if (dp[N][j] == true)
+            {
+                diff = sum-2*j;
+                break;
+            }
+        }
+        return diff;
+}