Given an array arr[] and two integers X and K. The task is to perform the following operation on array K times:
- Sort the array.
- XOR every alternate element of the sorted array with X i.e. arr[0], arr[2], arr[4], …
After repeating the above steps K times, print the maximum and the minimum element in the modified array.
Examples:
Input: arr[] = {9, 7, 11, 15, 5}, K = 1, X = 2
Output: 7 13
Since the operations has to be performed only once,
the sorted array will be {5, 7, 9, 11, 15}
Now, apply xor with 2 on alternate elements i.e. 5, 9 and 15.
{5 ^ 2, 7, 9 ^ 2, 11, 15 ^ 2} which is equal to
{7, 7, 11, 11, 13}
Input: arr[] = {605, 986}, K = 548, X = 569
Output: 605 986
Approach: Instead of sorting the array in every iteration, a frequency array can be maintained that will store the frequency of each element in the array. Traversing from 1 to maximum element in the array, the elements can be processed in the sorted order and after every operation, the frequency of the same elements can be adjusted when the alternate elements get XORed with the given integer. Check the programming implementation for more details.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 100000// Function to find the maximum and the// minimum elements from the array after// performing the given operation k timesvoid xorOnSortedArray(int arr[], int n, int k, int x){ // To store the current sequence of elements int arr1[MAX + 1] = { 0 }; // To store the next sequence of elements // after xoring with current elements int arr2[MAX + 1] = { 0 }; int xor_val[MAX + 1]; // Store the frequency of elements of arr[] in arr1[] for (int i = 0; i < n; i++) arr1[arr[i]]++; // Storing all precomputed XOR values so that // we don't have to do it again and again // as XOR is a costly operation for (int i = 0; i <= MAX; i++) xor_val[i] = i ^ x; // Perform the operations k times while (k--) { // The value of count decide on how many elements // we have to apply XOR operation int count = 0; for (int i = 0; i <= MAX; i++) { int store = arr1[i]; // If current element is present in // the array to be modified if (arr1[i] > 0) { // Suppose i = m and arr1[i] = num, it means // 'm' appears 'num' times // If the count is even we have to perform // XOR operation on alternate 'm' starting // from the 0th index because count is even // and we have to perform XOR operations // starting with initial 'm' // Hence there will be ceil(num/2) operations on // 'm' that will change 'm' to xor_val[m] i.e. m^x if (count % 2 == 0) { int div = ceil((float)arr1[i] / 2); // Decrease the frequency of 'm' from arr1[] arr1[i] = arr1[i] - div; // Increase the frequency of 'm^x' in arr2[] arr2[xor_val[i]] += div; } // If the count is odd we have to perform // XOR operation on alternate 'm' starting // from the 1st index because count is odd // and we have to leave the 0th 'm' // Hence there will be (num/2) XOR operations on // 'm' that will change 'm' to xor_val[m] i.e. m^x else if (count % 2 != 0) { int div = arr1[i] / 2; arr1[i] = arr1[i] - div; arr2[xor_val[i]] += div; } } // Updating the count by frequency of // the current elements as we have // processed that many elements count = count + store; } // Updating arr1[] which will now store the // next sequence of elements // At this time, arr1[] stores the remaining // 'm' on which XOR was not performed and // arr2[] stores the frequency of 'm^x' i.e. // those 'm' on which operation was performed // Updating arr1[] with frequency of remaining // 'm' & frequency of 'm^x' from arr2[] // With help of arr2[], we prevent sorting of // the array again and again for (int i = 0; i <= MAX; i++) { arr1[i] = arr1[i] + arr2[i]; // Resetting arr2[] for next iteration arr2[i] = 0; } } // Finding the maximum and the minimum element // from the modified array after the operations int min = INT_MAX; int max = INT_MIN; for (int i = 0; i <= MAX; i++) { if (arr1[i] > 0) { if (min > i) min = i; if (max < i) max = i; } } // Printing the max and the min element cout << min << " " << max << endl;}// Driver codeint main(){ int arr[] = { 605, 986 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 548, x = 569; xorOnSortedArray(arr, n, k, x); return 0;} |
Java
// Java implementation of the approach class GFG{ static int MAX = 100000; // Function to find the maximum and the // minimum elements from the array after // performing the given operation k times public static void xorOnSortedArray(int[] arr, int n, int k, int x) { // To store the current sequence of elements int[] arr1 = new int[MAX + 1]; // To store the next sequence of elements // after xoring with current elements int[] arr2 = new int[MAX + 1]; int[] xor_val = new int[MAX + 1]; // Store the frequency of elements // of arr[] in arr1[] for (int i = 0; i < n; i++) arr1[arr[i]]++; // Storing all precomputed XOR values so that // we don't have to do it again and again // as XOR is a costly operation for (int i = 0; i <= MAX; i++) xor_val[i] = i ^ x; // Perform the operations k times while (k-- > 0) { // The value of count decide on how many elements // we have to apply XOR operation int count = 0; for (int i = 0; i <= MAX; i++) { int store = arr1[i]; // If current element is present in // the array to be modified if (arr1[i] > 0) { // Suppose i = m and arr1[i] = num, it means // 'm' appears 'num' times // If the count is even we have to perform // XOR operation on alternate 'm' starting // from the 0th index because count is even // and we have to perform XOR operations // starting with initial 'm' // Hence there will be ceil(num/2) operations on // 'm' that will change 'm' to xor_val[m] i.e. m^x if (count % 2 == 0) { int div = (int) Math.ceil(arr1[i] / 2); // Decrease the frequency of 'm' from arr1[] arr1[i] = arr1[i] - div; // Increase the frequency of 'm^x' in arr2[] arr2[xor_val[i]] += div; } // If the count is odd we have to perform // XOR operation on alternate 'm' starting // from the 1st index because count is odd // and we have to leave the 0th 'm' // Hence there will be (num/2) XOR operations on // 'm' that will change 'm' to xor_val[m] i.e. m^x else if (count % 2 != 0) { int div = arr1[i] / 2; arr1[i] = arr1[i] - div; arr2[xor_val[i]] += div; } } // Updating the count by frequency of // the current elements as we have // processed that many elements count = count + store; } // Updating arr1[] which will now store the // next sequence of elements // At this time, arr1[] stores the remaining // 'm' on which XOR was not performed and // arr2[] stores the frequency of 'm^x' i.e. // those 'm' on which operation was performed // Updating arr1[] with frequency of remaining // 'm' & frequency of 'm^x' from arr2[] // With help of arr2[], we prevent sorting of // the array again and again for (int i = 0; i <= MAX; i++) { arr1[i] = arr1[i] + arr2[i]; // Resetting arr2[] for next iteration arr2[i] = 0; } } // Finding the maximum and the minimum element // from the modified array after the operations int min = Integer.MAX_VALUE; int max = Integer.MIN_VALUE; for (int i = 0; i <= MAX; i++) { if (arr1[i] > 0) { if (min > i) min = i; if (max < i) max = i; } } // Printing the max and the min element System.out.println(min + " " + max); } // Driver code public static void main(String[] args) { int[] arr = { 605, 986 }; int n = arr.length; int k = 548, x = 569; xorOnSortedArray(arr, n, k, x); }}// This code is contributed by// sanjeev2552 |
Python 3
# Python 3 implementation of the approachMAX = 10000import sysfrom math import ceil, floor# Function to find the maximum and the# minimum elements from the array after# performing the given operation k timesdef xorOnSortedArray(arr, n, k, x): # To store the current sequence of elements arr1 = [0 for i in range(MAX + 1)] # To store the next sequence of elements # after xoring with current elements arr2 = [0 for i in range(MAX + 1)] xor_val = [0 for i in range(MAX + 1)] # Store the frequency of elements # of arr[] in arr1[] for i in range(n): arr1[arr[i]] += 1 # Storing all precomputed XOR values # so that we don't have to do it # again and again as XOR is a costly operation for i in range(MAX + 1): xor_val[i] = i ^ x # Perform the operations k times while (k > 0): k -= 1 # The value of count decide on # how many elements we have to # apply XOR operation count = 0 for i in range(MAX + 1): store = arr1[i] # If current element is present in # the array to be modified if (arr1[i] > 0): # Suppose i = m and arr1[i] = num, # it means 'm' appears 'num' times # If the count is even we have to # perform XOR operation on alternate # 'm' starting from the 0th index because # count is even and we have to perform # XOR operations starting with initial 'm' # Hence there will be ceil(num/2) # operations on 'm' that will change # 'm' to xor_val[m] i.e. m^x if (count % 2 == 0): div = arr1[i] // 2 + 1 # Decrease the frequency of # 'm' from arr1[] arr1[i] = arr1[i] - div # Increase the frequency of # 'm^x' in arr2[] arr2[xor_val[i]] += div # If the count is odd we have to perform # XOR operation on alternate 'm' starting # from the 1st index because count is odd # and we have to leave the 0th 'm' # Hence there will be (num/2) XOR operations on # 'm' that will change 'm' to xor_val[m] i.e. m^x elif (count % 2 != 0): div = arr1[i] // 2 arr1[i] = arr1[i] - div arr2[xor_val[i]] += div # Updating the count by frequency of # the current elements as we have # processed that many elements count = count + store # Updating arr1[] which will now store the # next sequence of elements # At this time, arr1[] stores the remaining # 'm' on which XOR was not performed and # arr2[] stores the frequency of 'm^x' i.e. # those 'm' on which operation was performed # Updating arr1[] with frequency of remaining # 'm' & frequency of 'm^x' from arr2[] # With help of arr2[], we prevent sorting of # the array again and again for i in range(MAX+1): arr1[i] = arr1[i] + arr2[i] # Resetting arr2[] for next iteration arr2[i] = 0 # Finding the maximum and the minimum element # from the modified array after the operations mn = sys.maxsize mx = -sys.maxsize-1 for i in range(MAX + 1): if (arr1[i] > 0): if (mn > i): mn = i if (mx < i): mx = i # Printing the max and the min element print(mn,mx)# Driver codeif __name__ == '__main__': arr = [605, 986] n = len(arr) k = 548 x = 569 xorOnSortedArray(arr, n, k, x) # This code is contributed by Surendra_Gangwar |
C#
// C# implementation of the approach using System; class GFG{ static int MAX = 100000; // Function to find the maximum and the // minimum elements from the array after // performing the given operation k times public static void xorOnSortedArray(int[] arr, int n, int k, int x) { // To store the current sequence of elements int[] arr1 = new int[MAX + 1]; // To store the next sequence of elements // after xoring with current elements int[] arr2 = new int[MAX + 1]; int[] xor_val = new int[MAX + 1]; // Store the frequency of elements // of arr[] in arr1[] for (int i = 0; i < n; i++) arr1[arr[i]]++; // Storing all precomputed XOR values so that // we don't have to do it again and again // as XOR is a costly operation for (int i = 0; i <= MAX; i++) xor_val[i] = i ^ x; // Perform the operations k times while (k-- > 0) { // The value of count decides on // how many elements we have to // apply XOR operation int count = 0; for (int i = 0; i <= MAX; i++) { int store = arr1[i]; // If current element is present in // the array to be modified if (arr1[i] > 0) { // Suppose i = m and arr1[i] = num, // it means 'm' appears 'num' times // If the count is even we have to perform // XOR operation on alternate 'm' starting // from the 0th index because count is even // and we have to perform XOR operations // starting with initial 'm' // Hence there will be ceil(num/2) operations on // 'm' that will change 'm' to xor_val[m] i.e. m^x if (count % 2 == 0) { int div = (int) Math.Ceiling((double)(arr1[i] / 2)); // Decrease the frequency of 'm' from arr1[] arr1[i] = arr1[i] - div; // Increase the frequency of 'm^x' in arr2[] arr2[xor_val[i]] += div; } // If the count is odd we have to perform // XOR operation on alternate 'm' starting // from the 1st index because count is odd // and we have to leave the 0th 'm' // Hence there will be (num/2) XOR operations on // 'm' that will change 'm' to xor_val[m] i.e. m^x else if (count % 2 != 0) { int div = arr1[i] / 2; arr1[i] = arr1[i] - div; arr2[xor_val[i]] += div; } } // Updating the count by frequency of // the current elements as we have // processed that many elements count = count + store; } // Updating arr1[] which will now store the // next sequence of elements // At this time, arr1[] stores the remaining // 'm' on which XOR was not performed and // arr2[] stores the frequency of 'm^x' i.e. // those 'm' on which operation was performed // Updating arr1[] with frequency of remaining // 'm' & frequency of 'm^x' from arr2[] // With help of arr2[], we prevent sorting of // the array again and again for (int i = 0; i <= MAX; i++) { arr1[i] = arr1[i] + arr2[i]; // Resetting arr2[] for next iteration arr2[i] = 0; } } // Finding the maximum and the minimum element // from the modified array after the operations int min = int.MaxValue; int max = int.MinValue; for (int i = 0; i <= MAX; i++) { if (arr1[i] > 0) { if (min > i) min = i; if (max < i) max = i; } } // Printing the max and the min element Console.WriteLine(min + " " + max); } // Driver code public static void Main(String[] args) { int[] arr = { 605, 986 }; int n = arr.Length; int k = 548, x = 569; xorOnSortedArray(arr, n, k, x); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approachconst MAX = 100000;// Function to find the maximum and the// minimum elements from the array after// performing the given operation k timesfunction xorOnSortedArray(arr, n, k, x){ // To store the current sequence of elements let arr1 = new Array(MAX + 1).fill(0); // To store the next sequence of elements // after xoring with current elements let arr2 = new Array(MAX + 1).fill(0); let xor_val = new Array(MAX + 1); // Store the frequency of elements of arr[] in arr1[] for (let i = 0; i < n; i++) arr1[arr[i]]++; // Storing all precomputed XOR values so that // we don't have to do it again and again // as XOR is a costly operation for (let i = 0; i <= MAX; i++) xor_val[i] = i ^ x; // Perform the operations k times while (k--) { // The value of count decide on how many elements // we have to apply XOR operation let count = 0; for (let i = 0; i <= MAX; i++) { let store = arr1[i]; // If current element is present in // the array to be modified if (arr1[i] > 0) { // Suppose i = m and arr1[i] = num, it means // 'm' appears 'num' times // If the count is even we have to perform // XOR operation on alternate 'm' starting // from the 0th index because count is even // and we have to perform XOR operations // starting with initial 'm' // Hence there will be ceil(num/2) operations on // 'm' that will change 'm' to xor_val[m] i.e. m^x if (count % 2 == 0) { let div = Math.ceil(arr1[i] / 2); // Decrease the frequency of 'm' from arr1[] arr1[i] = arr1[i] - div; // Increase the frequency of 'm^x' in arr2[] arr2[xor_val[i]] += div; } // If the count is odd we have to perform // XOR operation on alternate 'm' starting // from the 1st index because count is odd // and we have to leave the 0th 'm' // Hence there will be (num/2) XOR operations on // 'm' that will change 'm' to xor_val[m] i.e. m^x else if (count % 2 != 0) { let div = parseInt(arr1[i] / 2); arr1[i] = arr1[i] - div; arr2[xor_val[i]] += div; } } // Updating the count by frequency of // the current elements as we have // processed that many elements count = count + store; } // Updating arr1[] which will now store the // next sequence of elements // At this time, arr1[] stores the remaining // 'm' on which XOR was not performed and // arr2[] stores the frequency of 'm^x' i.e. // those 'm' on which operation was performed // Updating arr1[] with frequency of remaining // 'm' & frequency of 'm^x' from arr2[] // With help of arr2[], we prevent sorting of // the array again and again for (let i = 0; i <= MAX; i++) { arr1[i] = arr1[i] + arr2[i]; // Resetting arr2[] for next iteration arr2[i] = 0; } } // Finding the maximum and the minimum element // from the modified array after the operations let min = Number.MAX_VALUE; let max = Number.MIN_VALUE; for (let i = 0; i <= MAX; i++) { if (arr1[i] > 0) { if (min > i) min = i; if (max < i) max = i; } } // Printing the max and the min element document.write(min + " " + max);}// Driver code let arr = [ 605, 986 ]; let n = arr.length; let k = 548, x = 569; xorOnSortedArray(arr, n, k, x);</script> |
Output:
605 986
Time Complexity: O(n + k * MAX)
Auxiliary Space: O(MAX)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
