Minimum insertions to make XOR of an Array equal to half of its sum

• Last Updated : 30 Apr, 2021

Given an array of positive integers, the task is to find the minimum number of insertions to be done in the array, to make the XOR of the array equal to half of its sum, i.e. 2 * Xor of all elements = Sum of all elements
Examples:

Input: arr[] = {1 2 3 4 5}
Output: 1 16
Explanation:
In the modified array {1 2 3 4 5 1 16},
Sum = 1 + 2 + 3 + 4 + 5 + 1 + 16 = 32
Xor = 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 1 ^ 16 = 16
And, 2 * 16 == 32
Thus, the condition 2 * Xor of all elements = Sum of all elements is satisfied.

Input: 7 11 3 25 51 32 9 29
Output: 17 184
Explanation:
In the modified array { 7 11 3 25 51 32 9 29 17 184}
Sum = 7 + 11 + 3 + 25 + 51 + 32 + 9 + 29 + 17 + 184 = 368
Xor = 7 ^ 11 ^ 3 ^ 25 ^ 51 ^ 32 ^ 9 ^ 29 ^ 17 ^ 184 = 184
And, 2 * 184 == 368
Thus, the condition 2 * Xor of all elements = Sum of all elements is satisfied.

Approach:
To solve the problem, we need to focus on the two basic properties of XOR:

• A xor A = 0
• A xor 0 = A

We need to follow the steps below to solve the problem:

• Calculate the Sum of all array elements(S) and the Xor of all elements (X). If S == 2*X, no change in array is required. Print -1 for this case.
• Otherwise, do the following:
1. If X = 0, just insert S into the array. Now, the XOR is S, and the sum is 2S.
2. Otherwise, Add X to the array to make the new Xor of the array equal to 0. Then, insert S+X in the array. Now, the Sum is 2(S+X) and the Xor is S+X

Below is the implementation of the above approach.

C++

 // C++ Program to make XOR of// of all array elements equal// to half of its sum// by minimum insertions #include using namespace std; // Function to make XOR of the// array equal to half of its sumint make_xor_half(vector& arr){    int sum = 0, xr = 0;     // Calculate the sum and    // Xor of all the elements    for (int a : arr) {        sum += a;        xr ^= a;    }     // If the required condition    // satisfies already, return    // the original array    if (2 * xr == sum)        return -1;     // If Xor is already zero,    // Insert sum    if (xr == 0) {        arr.push_back(sum);        return 1;    }     // Otherwise, insert xr    // and insert sum + xr    arr.push_back(xr);    arr.push_back(sum + xr);    return 2;} // Driver Codeint main(){     int N = 7;    vector nums        = { 3, 4, 7, 1, 2, 5, 6 };     int count = make_xor_half(nums);     if (count == -1)        cout << "-1" << endl;    else if (count == 1)        cout << nums[N] << endl;    else        cout << nums[N] << " "             << nums[N + 1] << endl;     return 0;}

Python3

 # Python3 program to make XOR of# of all array elements equal to # half of its sum by minimum # insertions # Function to make XOR of the# array equal to half of its sumdef make_xor_half(arr):     sum = 0; xr = 0;     # Calculate the sum and    # Xor of all the elements    for a in arr:        sum += a;        xr ^= a;     # If the required condition    # satisfies already, return    # the original array    if (2 * xr == sum):        return -1;     # If Xor is already zero,    # Insert sum    if (xr == 0):        arr.append(sum);        return 1;     # Otherwise, insert xr    # and insert sum + xr    arr.append(xr);    arr.append(sum + xr);    return 2; # Driver codeif __name__ == "__main__":     N = 7;    nums = [ 3, 4, 7, 1, 2, 5, 6 ];    count = make_xor_half(nums);     if (count == -1):        print("-1");             elif (count == 1):        print(nums[N]);             else:        print(nums[N], nums[N + 1]); # This code is contributed by AnkitRai01

Java

 // Java program to make XOR of all// array elements equal to half// of its sum by minimum insertionsimport java.util.*; class GFG{   // Function to make XOR of the// array equal to half of its sumstatic int make_xor_half(ArrayList arr){  int sum = 0, xr = 0;   // Calculate the sum and  // Xor of all the elements  for (int i = 0;           i < arr.size(); i++)   {    int a = arr.get(i);    sum += a;    xr ^= a;  }   // If the required condition  // satisfies already, return  // the original array  if (2 * xr == sum)    return -1;   // If Xor is already zero,  // Insert sum  if (xr == 0)  {    arr.add(sum);    return 1;  }   // Otherwise, insert xr  // and insert sum + xr  arr.add(xr);  arr.add(sum + xr);  return 2;} // Driver codepublic static void main(String[] args){  int N = 7;  ArrayList nums =  new ArrayList(      Arrays.asList(3, 4, 7,                    1, 2, 5, 6));   int count = make_xor_half(nums);   if (count == -1)    System.out.print(-1 + "\n");  else if (count == 1)    System.out.print(nums.get(N) + "\n");  else    System.out.print(nums.get(N) + " " +                     nums.get(N + 1) + "\n");}} // This code is contributed by Chitranayal

C#

 // C# program to make XOR of all// array elements equal to half// of its sum by minimum insertionsusing System;using System.Collections;using System.Collections.Generic; class GFG{   // Function to make XOR of the// array equal to half of its sumstatic int make_xor_half(ArrayList arr){    int sum = 0, xr = 0;      // Calculate the sum and    // Xor of all the elements    foreach(int a in arr)    {        sum += a;        xr ^= a;    }      // If the required condition    // satisfies already, return    // the original array    if (2 * xr == sum)        return -1;      // If Xor is already zero,    // Insert sum    if (xr == 0)    {        arr.Add(sum);        return 1;    }      // Otherwise, insert xr    // and insert sum + xr    arr.Add(xr);    arr.Add(sum + xr);    return 2;}   // Driver codepublic static void Main(string[] args){    int N = 7;    ArrayList nums = new ArrayList(){ 3, 4, 7, 1,                                      2, 5, 6 };      int count = make_xor_half(nums);      if (count == -1)        Console.Write(-1 + "\n");    else if (count == 1)        Console.Write(nums[N] + "\n");    else        Console.Write(nums[N] + " " +                      nums[N + 1] + "\n");}} // This code is contributed by rutvik_56

Javascript


Output:
28

Time Complexity: O(N) where N is the size of the array.

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