# Pair of arrays with equal sum after removing exactly one element from each

• Difficulty Level : Medium
• Last Updated : 02 Jul, 2018

Given K arrays of different size. The task is to check if there exist any two arrays which have the same sum of elements after removing exactly one element from each of them. (Any element can be removed, but exactly one has to be removed). Print the indices of the array and the index of the removed elements if such pairs exist. If there are multiple pairs, print any one of them. If no such pairs exist, print -1.

Examples:

Input: k = 3
a1 = {8, 1, 4, 7, 1}
a2 = {10, 10}
a3 = {1, 3, 4, 7, 3, 2, 2}
Output: Array 1, index 4
Array 3, index 5
sum of Array 1{8, 1, 4, 7, 1} without index 4 is 20.
sum of Array 3{1, 3, 4, 7, 3, 2, 2} without index 5 is 20.

Input: k = 4
a1 = {2, 4, 6, 6}
a2 = {1, 2, 4, 8, 16}
a3 = {1, 3, 8}
a4 = {1, 4, 16, 64}
Output: -1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Brute Force:
For every pair of arrays, for each element, find the sum excluding that element and compare it with the sum excluding each element one by one in the second array of the chosen pair.
(Here maxl denotes the maximum length of an array in the set).

Time Complexity : Space Complexity : Efficient Approach: Precompute all possible values of the sum obtained by removing one element from each of the arrays. Store the array index and element index which is removed with the computed sum. When these values are arranged in increasing order, it can easily be seen that if a solution exists, then both the sum values must be adjacent to the new arrangement. When two adjacent sum values are same, check if they belong to different arrays. If they do, print the array number and index of the element removed. If no such sum value is found, then no such pairs exist.

Below is the implementation of the above approach:

 // C++program to print the pair of arrays  // whose sum are equal after removing  // exactly one element from each  #include  using namespace std;     // Function to print the pair of array and index  // of element to be removed to make sum equal  void printPair(vector<int> a[], int k)  {         // stores the sum removing one element,      // array number and index of removed element      vector > > ans;         // traverse in every array      for (int i = 0; i < k; i++) {             // length of array          int l = a[i].size();             int sum = 0;             // compute total sum of array          for (int j = 0; j < l; j++) {              sum = sum + a[i][j];          }             // remove each element once and insert sum in          // ans vector along with index          for (int j = 0; j < l; j++) {              ans.push_back({ sum - a[i][j], { i + 1, j } });          }      }         // sort the ans vector so that      // same sum values after removing      // a single element comes together      sort(ans.begin(), ans.end());         bool flag = false;         // iterate and check if any adjacent sum are equal      for (int p = 1; p < ans.size(); p++) {             // check if the adjacent sum belong to different array          // if the adjacent sum is equal          if (ans[p - 1].first == ans[p].first              && ans[p - 1].second.first != ans[p].second.first) {                 // first array number              int ax = ans[p - 1].second.first;                 // element's index removed from first array              int aidx = ans[p - 1].second.second;                 // second  array number              int bx = ans[p].second.first;                 // element's index removed from second array              int bidx = ans[p].second.second;                 cout << "Array " << ax << ", index " << aidx << "\n";              cout << "Array " << bx << ", index " << bidx << "\n";                 flag = true;              break;          }      }         // If no pairs are found      if (!flag)          cout << "No special pair exists\n";  }     // Driver Code  int main()  {      // k sets of array      vector<int> a[] = {          { 8, 1, 4, 7, 1 },          { 10, 10 },          { 1, 3, 4, 7, 3, 2, 2 }      };      int k = sizeof(a) / sizeof(a);         // Calling Function to print the pairs if any      printPair(a, k);         return 0;  }

Output:

Array 1, index 4
Array 3, index 5


Time Complexity : , or simply, Space Complexity : , or simply, My Personal Notes arrow_drop_up
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