Question

Program done in PROLOG

Print equal sum sets of array (Partition problem)

Given an array arr[]. Determine whether it is possible to split the array into two sets such that the sum of elements in both the sets is equal. If it is possible, then print both the sets. If it is not possible then output -1.

Answer 1

Code is as follows-

 // CPP program to print equal sum sets of array. #include <bits/stdc++.h> using namespace std; // Function to print equal sum // sets of array. void printEqualSumSets(int arr[], int n) {     int i, currSum;                 // Finding sum of array elements        int sum = accumulate(arr, arr+n, 0);    // Check sum is even or odd. If odd     // then array cannot be partitioned.    // Print -1 and return.         if (sum & 1) {              cout << "-1";             return;         }       // Divide sum by 2 to find      // sum of two possible subsets.         int k = sum >> 1;         // Boolean DP table to store result     // of states.   // dp[i][j] = true if there is a        // subset of elements in first i elements       // of array that has sum equal to j.    bool dp[n + 1][k + 1];  // If number of elements are zero, then         // no sum can be obtained.      for (i = 1; i <= k; i++)             dp[0][i] = false;       // Sum 0 can be obtained by not selecting       // any element.         for (i = 0; i <= n; i++)             dp[i][0] = true;        // Fill the DP table in bottom up manner.       for (i = 1; i <= n; i++) {           for (currSum = 1; currSum <= k; currSum++) {                         // Excluding current element.                   dp[i][currSum] = dp[i - 1][currSum];                    // Including current element                    if (arr[i - 1] <= currSum)                           dp[i][currSum] = dp[i][currSum] |                               dp[i - 1][currSum - arr[i - 1]];                }       }       // Required sets set1 and set2.         vector<int> set1, set2;   // If partition is not possible print   // -1 and return.       if (!dp[n][k]) {                cout << "-1\n";           return;         }       // Start from last element in dp table.         i = n;  currSum = k;    while (i > 0 && currSum >= 0) {           // If current element does not          // contribute to k, then it belongs             // to set 2.            if (dp[i - 1][currSum]) {                       i--;                    set2.push_back(arr[i]);                 }               // If current element contribute                // to k then it belongs to set 1.               else if (dp[i - 1][currSum - arr[i - 1]]) {                     i--;                    currSum -= arr[i];                      set1.push_back(arr[i]);                 }       }       // Print elements of both the sets.     cout << "Set 1 elements: ";       for (i = 0; i < set1.size(); i++)            cout << set1[i] << " ";     cout << "\nSet 2 elements: ";     for (i = 0; i < set2.size(); i++)            cout << set2[i] << " ";      } // Driver program. int main() {      int arr[] = { 5, 5, 1, 11 };    int n = sizeof(arr) / sizeof(arr[0]);   printEqualSumSets(arr, n);      return 0; } 

Output-


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