Given an array arr[] consisting of N positive integers and an integer K, which represents the maximum number that can be added to the array elements. The task is to maximize the length of the longest possible subarray of equal elements by adding at most K.
Examples:
Input: arr[] = {3, 0, 2, 2, 1}, k = 3
Output: 4
Explanation:
Step 1: Adding 2 to arr[1] modifies array to {3, 2, 2, 2, 1}
Step 2: Adding 1 to arr[4] modifies array to {3, 2, 2, 2, 2}
Therefore, answer will be 4 ({arr[1], …, arr[4]}).Input: arr[] = {1, 1, 1}, k = 7
Output: 3
Explanation:
All array elements are already equal. Therefore, the length is 3.
Approach: Follow the steps below to solve the problem:
- Sort the array arr[]. Then, use Binary Search to pick a possible value for the maximum indices having the same element.
- For each picked_value, use the Sliding Window technique to check if it is possible to make all elements equal for any subarray of size picked_value.
- Finally, print the longest possible length of subarray obtained.
Below is the implementation for the above approach:
C++14
// C++14 program for above approach #include <bits/stdc++.h>using namespace std;// Function to check if a subarray of// length len consisting of equal elements// can be obtained or notbool check(vector<int> pSum, int len, int k, vector<int> a){ // Sliding window int i = 0; int j = len; while (j <= a.size()) { // Last element of the sliding window // will be having the max size in the // current window int maxSize = a[j - 1]; int totalNumbers = maxSize * len; // The current number of element in all // indices of the current sliding window int currNumbers = pSum[j] - pSum[i]; // If the current number of the window, // added to k exceeds totalNumbers if (currNumbers + k >= totalNumbers) { return true; } else { i++; j++; } } return false;}// Function to find the maximum number of// indices having equal elements after// adding at most k numbersint maxEqualIdx(vector<int> arr, int k){ // Sort the array in // ascending order sort(arr.begin(), arr.end()); // Make prefix sum array vector<int> prefixSum(arr.size()); prefixSum[1] = arr[0]; for(int i = 1; i < prefixSum.size() - 1; ++i) { prefixSum[i + 1] = prefixSum[i] + arr[i]; } // Initialize variables int max = arr.size(); int min = 1; int ans = 1; while (min <= max) { // Update mid int mid = (max + min) / 2; // Check if any subarray // can be obtained of length // mid having equal elements if (check(prefixSum, mid, k, arr)) { ans = mid; min = mid + 1; } else { // Decrease max to mid max = mid - 1; } } return ans;}// Driver Codeint main(){ vector<int> arr = { 1, 1, 1 }; int k = 7; // Function call cout << (maxEqualIdx(arr, k));}// This code is contributed by mohit kumar 29 |
Java
// Java program for above approachimport java.util.*;class GFG { // Function to find the maximum number of // indices having equal elements after // adding at most k numbers public static int maxEqualIdx(int[] arr, int k) { // Sort the array in // ascending order Arrays.sort(arr); // Make prefix sum array int[] prefixSum = new int[arr.length + 1]; prefixSum[1] = arr[0]; for (int i = 1; i < prefixSum.length - 1; ++i) { prefixSum[i + 1] = prefixSum[i] + arr[i]; } // Initialize variables int max = arr.length; int min = 1; int ans = 1; while (min <= max) { // Update mid int mid = (max + min) / 2; // Check if any subarray // can be obtained of length // mid having equal elements if (check(prefixSum, mid, k, arr)) { ans = mid; min = mid + 1; } else { // Decrease max to mid max = mid - 1; } } return ans; } // Function to check if a subarray of // length len consisting of equal elements // can be obtained or not public static boolean check(int[] pSum, int len, int k, int[] a) { // Sliding window int i = 0; int j = len; while (j <= a.length) { // Last element of the sliding window // will be having the max size in the // current window int maxSize = a[j - 1]; int totalNumbers = maxSize * len; // The current number of element in all // indices of the current sliding window int currNumbers = pSum[j] - pSum[i]; // If the current number of the window, // added to k exceeds totalNumbers if (currNumbers + k >= totalNumbers) { return true; } else { i++; j++; } } return false; } // Driver Code public static void main(String[] args) { int[] arr = { 1, 1, 1 }; int k = 7; // Function call System.out.println(maxEqualIdx(arr, k)); }} |
Python3
# Python3 program for above approach # Function to find the maximum number of# indices having equal elements after# adding at most k numbersdef maxEqualIdx(arr, k): # Sort the array in # ascending order arr.sort() # Make prefix sum array prefixSum = [0] * (len(arr) + 1) prefixSum[1] = arr[0] for i in range(1, len(prefixSum) - 1 ,1): prefixSum[i + 1] = prefixSum[i] + arr[i] # Initialize variables max = len(arr) min = 1 ans = 1 while (min <= max): # Update mid mid = (max + min) // 2 # Check if any subarray # can be obtained of length # mid having equal elements if (check(prefixSum, mid, k, arr)): ans = mid min = mid + 1 else: # Decrease max to mid max = mid - 1 return ans# Function to check if a subarray of# length len consisting of equal elements# can be obtained or notdef check(pSum, lenn, k, a): # Sliding window i = 0 j = lenn while (j <= len(a)): # Last element of the sliding window # will be having the max size in the # current window maxSize = a[j - 1] totalNumbers = maxSize * lenn # The current number of element in all # indices of the current sliding window currNumbers = pSum[j] - pSum[i] # If the current number of the window, # added to k exceeds totalNumbers if (currNumbers + k >= totalNumbers): return True else: i += 1 j += 1 return False# Driver Codearr = [ 1, 1, 1 ] k = 7 # Function callprint(maxEqualIdx(arr, k))# This code is contributed by code_hunt |
C#
// C# program for // the above approachusing System;class GFG{// Function to find the maximum number of// indices having equal elements after// adding at most k numberspublic static int maxEqualIdx(int[] arr, int k){ // Sort the array in // ascending order Array.Sort(arr); // Make prefix sum array int[] prefixSum = new int[arr.Length + 1]; prefixSum[1] = arr[0]; for (int i = 1; i < prefixSum.Length - 1; ++i) { prefixSum[i + 1] = prefixSum[i] + arr[i]; } // Initialize variables int max = arr.Length; int min = 1; int ans = 1; while (min <= max) { // Update mid int mid = (max + min) / 2; // Check if any subarray // can be obtained of length // mid having equal elements if (check(prefixSum, mid, k, arr)) { ans = mid; min = mid + 1; } else { // Decrease max to mid max = mid - 1; } } return ans;}// Function to check if a subarray of// length len consisting of equal elements// can be obtained or notpublic static bool check(int[] pSum, int len, int k, int[] a){ // Sliding window int i = 0; int j = len; while (j <= a.Length) { // Last element of the sliding window // will be having the max size in the // current window int maxSize = a[j - 1]; int totalNumbers = maxSize * len; // The current number of element in all // indices of the current sliding window int currNumbers = pSum[j] - pSum[i]; // If the current number of the window, // added to k exceeds totalNumbers if (currNumbers + k >= totalNumbers) { return true; } else { i++; j++; } } return false;}// Driver Codepublic static void Main(String[] args){ int[] arr = {1, 1, 1}; int k = 7; // Function call Console.WriteLine(maxEqualIdx(arr, k));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program to implement// the above approach // Function to find the maximum number of // indices having equal elements after // adding at most k numbers function maxEqualIdx(arr, k) { // Sort the array in // ascending order arr.sort(); // Make prefix sum array let prefixSum = new Array(arr.length + 1).fill(0); prefixSum[1] = arr[0]; for (let i = 1; i < prefixSum.length - 1; ++i) { prefixSum[i + 1] = prefixSum[i] + arr[i]; } // Initialize variables let max = arr.length; let min = 1; let ans = 1; while (min <= max) { // Update mid let mid = Math.floor((max + min) / 2); // Check if any subarray // can be obtained of length // mid having equal elements if (check(prefixSum, mid, k, arr)) { ans = mid; min = mid + 1; } else { // Decrease max to mid max = mid - 1; } } return ans; } // Function to check if a subarray of // length len consisting of equal elements // can be obtained or not function check(pSum, len, k, a) { // Sliding window let i = 0; let j = len; while (j <= a.length) { // Last element of the sliding window // will be having the max size in the // current window let maxSize = a[j - 1]; let totalNumbers = maxSize * len; // The current number of element in all // indices of the current sliding window let currNumbers = pSum[j] - pSum[i]; // If the current number of the window, // added to k exceeds totalNumbers if (currNumbers + k >= totalNumbers) { return true; } else { i++; j++; } } return false; }// Driver Code let arr = [ 1, 1, 1 ]; let k = 7; // Function call document.write(maxEqualIdx(arr, k));</script> |
Output:
3
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
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