Minimizing array sum by applying XOR operation on all elements of the array
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 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 = 26Input: 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 array corresponding to position of each bit in element in array.
- Now, Traverse the array and check whether 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 sum int 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 code int 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 approach 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], 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 approach MAX = 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 approach using 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 */ |
Output:
26
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