Given an array arr[] of N integer elements, the task is to choose an element X and apply XOR operation on every element of the array with X such that the array sum is minimized.
Input: arr[] = {3, 5, 7, 11, 15}
Output: 26
Binary representation of the array elements are {0011, 0101, 0111, 1011, 1111}
We take xor of every element with 7 in order to minimize the sum.
3 XOR 7 = 0100 (4)
5 XOR 7 = 0010 (2)
7 XOR 7 = 0000 (0)
11 XOR 7 = 1100 (12)
15 XOR 7 = 1000 (8)
Sum = 4 + 2 + 0 + 12 + 8 = 26
Input: arr[] = {1, 2, 3, 4, 5}
Output: 14
Approach: The task is to find the element X with which we have to take xor of each element.
- Convert each number into binary form and update the frequency of bit (0 or 1) in an array corresponding to the position of each bit in the element in the array.
- Now, Traverse the array and check whether the element at index is more than n/2 (for ‘n’ elements, we check whether the set bit appears more than n/2 at index), and subsequently, we obtain element ‘X’
- Now, take xor of ‘X’ with all the elements and return the sum.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;const int MAX = 25;// Function to return the minimized sumint getMinSum(int arr[], int n){ int bits_count[MAX], max_bit = 0, sum = 0, ans = 0; memset(bits_count, 0, sizeof(bits_count)); // To store the frequency // of bit in every element for (int d = 0; d < n; d++) { int e = arr[d], f = 0; while (e > 0) { int rem = e % 2; e = e / 2; if (rem == 1) { bits_count[f] += rem; } f++; } max_bit = max(max_bit, f); } // Finding element X for (int d = 0; d < max_bit; d++) { int temp = pow(2, d); if (bits_count[d] > n / 2) ans = ans + temp; } // Taking XOR of elements and finding sum for (int d = 0; d < n; d++) { arr[d] = arr[d] ^ ans; sum = sum + arr[d]; } return sum;}// Driver codeint main(){ int arr[] = { 3, 5, 7, 11, 15 }; int n = sizeof(arr) / sizeof(arr[0]); cout << getMinSum(arr, n); return 0;} |
Java
// Java implementation of the approachclass GFG { static int MAX = 25; // Function to return the minimized sum static int getMinSum(int arr[], int n) { int bits_count[] = new int[MAX], max_bit = 0, sum = 0, ans = 0; // To store the frequency // of bit in every element for (int d = 0; d < n; d++) { int e = arr[d], f = 0; while (e > 0) { int rem = e % 2; e = e / 2; if (rem == 1) { bits_count[f] += rem; } f++; } max_bit = Math.max(max_bit, f); } // Finding element X for (int d = 0; d < max_bit; d++) { int temp = (int)Math.pow(2, d); if (bits_count[d] > n / 2) ans = ans + temp; } // Taking XOR of elements and finding sum for (int d = 0; d < n; d++) { arr[d] = arr[d] ^ ans; sum = sum + arr[d]; } return sum; } // Driver code public static void main(String[] args) { int arr[] = { 3, 5, 7, 11, 15 }; int n = arr.length; System.out.println(getMinSum(arr, n)); }}// This code has been contributed by 29AjayKumar |
Python3
# Python3 implementation of the approachMAX = 25; # Function to return the minimized sum def getMinSum(arr, n) : bits_count = [0]* MAX max_bit = 0; sum = 0; ans = 0; # To store the frequency # of bit in every element for d in range(n) : e = arr[d]; f = 0; while (e > 0) : rem = e % 2; e = e // 2; if (rem == 1) : bits_count[f] += rem; f += 1 max_bit = max(max_bit, f); # Finding element X for d in range(max_bit) : temp = pow(2, d); if (bits_count[d] > n // 2) : ans = ans + temp; # Taking XOR of elements and finding sum for d in range(n) : arr[d] = arr[d] ^ ans; sum = sum + arr[d]; return sum # Driver code if __name__ == "__main__" : arr = [ 3, 5, 7, 11, 15 ]; n = len(arr); print(getMinSum(arr, n))# This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;class GFG { static int MAX = 25; // Function to return the minimized sum static int getMinSum(int[] arr, int n) { int[] bits_count = new int[MAX]; int max_bit = 0, sum = 0, ans = 0; // To store the frequency // of bit in every element for (int d = 0; d < n; d++) { int e = arr[d], f = 0; while (e > 0) { int rem = e % 2; e = e / 2; if (rem == 1) { bits_count[f] += rem; } f++; } max_bit = Math.Max(max_bit, f); } // Finding element X for (int d = 0; d < max_bit; d++) { int temp = (int)Math.Pow(2, d); if (bits_count[d] > n / 2) ans = ans + temp; } // Taking XOR of elements and finding sum for (int d = 0; d < n; d++) { arr[d] = arr[d] ^ ans; sum = sum + arr[d]; } return sum; } // Driver code public static void Main(String[] args) { int[] arr = { 3, 5, 7, 11, 15 }; int n = arr.Length; Console.WriteLine(getMinSum(arr, n)); }}/* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// Javascript implementation of the approachconst MAX = 25;// Function to return the minimized sumfunction getMinSum(arr, n){ let bits_count = new Array(MAX).fill(0), max_bit = 0, sum = 0, ans = 0; // To store the frequency // of bit in every element for (let d = 0; d < n; d++) { let e = arr[d], f = 0; while (e > 0) { let rem = e % 2; e = parseInt(e / 2); if (rem == 1) { bits_count[f] += rem; } f++; } max_bit = Math.max(max_bit, f); } // Finding element X for (let d = 0; d < max_bit; d++) { let temp = Math.pow(2, d); if (bits_count[d] > parseInt(n / 2)) ans = ans + temp; } // Taking XOR of elements and finding sum for (let d = 0; d < n; d++) { arr[d] = arr[d] ^ ans; sum = sum + arr[d]; } return sum;}// Driver code let arr = [ 3, 5, 7, 11, 15 ]; let n = arr.length; document.write(getMinSum(arr, n));</script> |
Output:
26
Time Complexity: O(N*log(max_element)), as we are using nested loops the outer loop traverses N times and the inner loop traverses log(max_element) times. In inner loop traversal, we are decrementing by floor division of 2 in each traversal equivalent to 1+1/2+1/4+….1/2max_element. Where N is the number of elements in the array and max_element is the maximum element present in the array.
Auxiliary Space: O(1), as we are not using any extra space.
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