Given an array arr[] of positive integers of size N and a positive integer K, the task is to find the maximum possible length of a subarray which can be made equal by adding some integer value to each element of the sub-array such that the sum of the added elements does not exceed K.
Examples:
Input: N = 5, arr[] = {1, 4, 9, 3, 6}, K = 9
Output: 3
Explanation:
{1, 4} : {1+3, 4} = {4, 4}
{4, 9} : {4+5, 9} = {9, 9}
{3, 6} : {3+3, 6} = {6, 6}
{9, 3, 6} : {9, 3+6, 6+3} = {9, 9, 9}
Hence, the maximum length of such a subarray is 3.Input: N = 6, arr[] = {2, 4, 7, 3, 8, 5}, K = 10
Output: 4
Approach: This problem can be solved by using dynamic programming.
- Initialize:
- dp[]: Stores the sum of elements that are added to the subarray.
- deque: Stores the indices of the maximum element for each subarray.
- pos: Index of the current position of the subarray.
- ans: Length of the maximum subarray.
- mx: Maximum element of a subarray
- pre: Previous index of the current subarray.
- Traverse the array and check if the deque is empty or not. If yes, then update the maximum element and the index of the maximum element along with the indices of pre and pos.
- Check if the currently added element is greater than K. If yes, then remove it from dp[] array and update the indices of pos and pre.
- Finally, update the maximum length of the valid sub-array.
Below is the implementation of the above approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to find maximum// possible length of subarrayint validSubArrLength(int arr[], int N, int K){ // Stores the sum of elements // that needs to be added to // the sub array int dp[N + 1]; // Stores the index of the // current position of subarray int pos = 0; // Stores the maximum // length of subarray. int ans = 0; // Maximum element from // each subarray length int mx = 0; // Previous index of the // current subarray of // maximum length int pre = 0; // Deque to store the indices // of maximum element of // each sub array deque<int> q; // For each array element, // find the maximum length of // required subarray for (int i = 0; i < N; i++) { // Traverse the deque and // update the index of // maximum element. while (!q.empty() && arr[q.back()] < arr[i]) q.pop_back(); q.push_back(i); // If it is first element // then update maximum // and dp[] if (i == 0) { mx = arr[i]; dp[i] = arr[i]; } // Else check if current // element exceeds max else if (mx <= arr[i]) { // Update max and dp[] dp[i] = dp[i - 1] + arr[i]; mx = arr[i]; } else { dp[i] = dp[i - 1] + arr[i]; } // Update the index of the // current maximum length // subarray if (pre == 0) pos = 0; else pos = pre - 1; // While current element // being added to dp[] array // exceeds K while ((i - pre + 1) * mx - (dp[i] - dp[pos]) > K && pre < i) { // Update index of // current position and // the previous position pos = pre; pre++; // Remove elements // from deque and // update the // maximum element while (!q.empty() && q.front() < pre && pre < i) { q.pop_front(); mx = arr[q.front()]; } } // Update the maximum length // of the required subarray. ans = max(ans, i - pre + 1); } return ans;}// Driver Programint main(){ int N = 6; int K = 8; int arr[] = { 2, 7, 1, 3, 4, 5 }; cout << validSubArrLength(arr, N, K); return 0;} |
Java
// Java code for the above approachimport java.util.*;class GFG{// Function to find maximum// possible length of subarraystatic int validSubArrLength(int arr[], int N, int K){ // Stores the sum of elements // that needs to be added to // the sub array int []dp = new int[N + 1]; // Stores the index of the // current position of subarray int pos = 0; // Stores the maximum // length of subarray. int ans = 0; // Maximum element from // each subarray length int mx = 0; // Previous index of the // current subarray of // maximum length int pre = 0; // Deque to store the indices // of maximum element of // each sub array Deque<Integer> q = new LinkedList<>(); // For each array element, // find the maximum length of // required subarray for(int i = 0; i < N; i++) { // Traverse the deque and // update the index of // maximum element. while (!q.isEmpty() && arr[q.getLast()] < arr[i]) q.removeLast(); q.add(i); // If it is first element // then update maximum // and dp[] if (i == 0) { mx = arr[i]; dp[i] = arr[i]; } // Else check if current // element exceeds max else if (mx <= arr[i]) { // Update max and dp[] dp[i] = dp[i - 1] + arr[i]; mx = arr[i]; } else { dp[i] = dp[i - 1] + arr[i]; } // Update the index of the // current maximum length // subarray if (pre == 0) pos = 0; else pos = pre - 1; // While current element // being added to dp[] array // exceeds K while ((i - pre + 1) * mx - (dp[i] - dp[pos]) > K && pre < i) { // Update index of // current position and // the previous position pos = pre; pre++; // Remove elements from // deque and update the // maximum element while (!q.isEmpty() && q.peek() < pre && pre < i) { q.removeFirst(); mx = arr[q.peek()]; } } // Update the maximum length // of the required subarray. ans = Math.max(ans, i - pre + 1); } return ans;}// Driver codepublic static void main(String[] args){ int N = 6; int K = 8; int arr[] = { 2, 7, 1, 3, 4, 5 }; System.out.print(validSubArrLength(arr, N, K));}}// This code is contributed by amal kumar choubey |
Python3
# Python3 code for the above approach# Function to find maximum# possible length of subarraydef validSubArrLength(arr, N, K): # Stores the sum of elements # that needs to be added to # the sub array dp = [0 for i in range(N + 1)] # Stores the index of the # current position of subarray pos = 0 # Stores the maximum # length of subarray. ans = 0 # Maximum element from # each subarray length mx = 0 # Previous index of the # current subarray of # maximum length pre = 0 # Deque to store the indices # of maximum element of # each sub array q = [] # For each array element, # find the maximum length of # required subarray for i in range(N): # Traverse the deque and # update the index of # maximum element. while (len(q) and arr[len(q) - 1] < arr[i]): q.remove(q[len(q) - 1]) q.append(i) # If it is first element # then update maximum # and dp[] if (i == 0): mx = arr[i] dp[i] = arr[i] # Else check if current # element exceeds max elif (mx <= arr[i]): # Update max and dp[] dp[i] = dp[i - 1] + arr[i] mx = arr[i] else: dp[i] = dp[i - 1] + arr[i] # Update the index of the # current maximum length # subarray if (pre == 0): pos = 0 else: pos = pre - 1 # While current element # being added to dp[] array # exceeds K while ((i - pre + 1) * mx - (dp[i] - dp[pos]) > K and pre < i): # Update index of # current position and # the previous position pos = pre pre += 1 # Remove elements # from deque and # update the # maximum element while (len(q) and q[0] < pre and pre < i): q.remove(q[0]) mx = arr[q[0]] # Update the maximum length # of the required subarray. ans = max(ans, i - pre + 1) return ans# Driver codeif __name__ == '__main__': N = 6 K = 8 arr = [ 2, 7, 1, 3, 4, 5 ] print(validSubArrLength(arr, N, K))# This code is contributed by ipg2016107 |
Javascript
<script>// Javascript code for the above approach// Function to find maximum// possible length of subarrayfunction validSubArrLength(arr, N, K){ // Stores the sum of elements // that needs to be added to // the sub array var dp = Array(N+1); // Stores the index of the // current position of subarray var pos = 0; // Stores the maximum // length of subarray. var ans = 0; // Maximum element from // each subarray length var mx = 0; // Previous index of the // current subarray of // maximum length var pre = 0; // Deque to store the indices // of maximum element of // each sub array var q = []; // For each array element, // find the maximum length of // required subarray for (var i = 0; i < N; i++) { // Traverse the deque and // update the index of // maximum element. while (q.length!=0 && arr[q[q.length-1]] < arr[i]) q.pop(); q.push(i); // If it is first element // then update maximum // and dp[] if (i == 0) { mx = arr[i]; dp[i] = arr[i]; } // Else check if current // element exceeds max else if (mx <= arr[i]) { // Update max and dp[] dp[i] = dp[i - 1] + arr[i]; mx = arr[i]; } else { dp[i] = dp[i - 1] + arr[i]; } // Update the index of the // current maximum length // subarray if (pre == 0) pos = 0; else pos = pre - 1; // While current element // being added to dp[] array // exceeds K while ((i - pre + 1) * mx - (dp[i] - dp[pos]) > K && pre < i) { // Update index of // current position and // the previous position pos = pre; pre++; // Remove elements // from deque and // update the // maximum element while (q.length!=0 && q[0] < pre && pre < i) { q.shift(); mx = arr[q[0]]; } } // Update the maximum length // of the required subarray. ans = Math.max(ans, i - pre + 1); } return ans;}// Driver Programvar N = 6;var K = 8;var arr = [2, 7, 1, 3, 4, 5];document.write( validSubArrLength(arr, N, K));</script> |
C#
// C# program to implement above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{ // Function to find maximum // possible length of subarray static int validSubArrLength(int[] arr, int N, int K) { // Stores the sum of elements // that needs to be added to // the sub array int[] dp = new int[N + 1]; // Stores the index of the // current position of subarray int pos = 0; // Stores the maximum // length of subarray. int ans = 0; // Maximum element from // each subarray length int mx = 0; // Previous index of the // current subarray of // maximum length int pre = 0; // Deque to store the indices // of maximum element of // each sub array List<int> q = new List<int>(); // For each array element, // find the maximum length of // required subarray for(int i = 0 ; i < N ; i++) { // Traverse the deque and // update the index of // maximum element. while (q.Count > 0 && arr[q[q.Count - 1]] < arr[i]){ q.RemoveAt(q.Count - 1); } q.Add(i); // If it is first element // then update maximum // and dp[] if (i == 0) { mx = arr[i]; dp[i] = arr[i]; } // Else check if current // element exceeds max else if (mx <= arr[i]) { // Update max and dp[] dp[i] = dp[i - 1] + arr[i]; mx = arr[i]; } else { dp[i] = dp[i - 1] + arr[i]; } // Update the index of the // current maximum length // subarray if (pre == 0) pos = 0; else pos = pre - 1; // While current element // being added to dp[] array // exceeds K while ((i - pre + 1) * mx - (dp[i] - dp[pos]) > K && pre < i) { // Update index of // current position and // the previous position pos = pre; pre++; // Remove elements from // deque and update the // maximum element while (q.Count > 0 && q[0] < pre && pre < i) { q.RemoveAt(0); mx = arr[q[0]]; } } // Update the maximum length // of the required subarray. ans = Math.Max(ans, i - pre + 1); } return ans; } // Driver code public static void Main(string[] args){ int N = 6; int K = 8; int[] arr = new int[]{ 2, 7, 1, 3, 4, 5 }; Console.WriteLine(validSubArrLength(arr, N, K)); }} |
Output:
4
Time Complexity: O(N2)
Auxiliary Space Complexity: O(N)
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