Given an array arr[] consisting of N positive integers and a positive integer K, the task is to find the maximum possible integer X, such that the absolute difference between any array element and X is at most K. If no such value of X exists, then print “-1”.
Examples:
Input: arr[] = {6, 4, 8, 5}, K = 2
Output: 6
Explanation: Considering X to be 6, the absolute difference between every array element and X(= 6) is at most K (= 2), as illustrated below:
- Absolute difference between arr[0](= 6) and X(= 6) = |6 – 6| = 0.
- Absolute difference between arr[1](= 4) and X(= 6) = |4 – 6| = 2.
- Absolute difference between arr[2](= 8) and X(= 6) = |8 – 6| = 2.
- Absolute difference between arr[3](= 5) and X(= 6) = |5 – 6| = 1.
Input: arr[] = {1, 2, 5}, K = 2
Output: 3
Approach: The given problem can be solved based on the following observations:
- Considering array elements to be arr[i], the value of |arr[i] – X| must be at most K.
- If arr[i] > X, then X ? (arr[i] – K). Otherwise, X ? (arr[i] + K).
- From the above two equations, the maximum value of X must be the sum of minimum value of arr[i] and K.
Follow the steps below to solve the problem:
- Find the smallest element of the given array arr[] in a variable, say S.
- Consider the maximum value of X as (S + K).
- Traverse the given array arr[] and if the value of the absolute difference of arr[i] and X is K, then update X to -1 and break out of the loop. Otherwise, continue the iteration.
- After completing the above steps, print the value of X as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find maximum value// of X such that |A[i] - X| ? Kint maximumNumber(int arr[], int N, int K){ // Stores the smallest array element int minimum = *min_element(arr, arr + N); // Store the possible value of X int ans = minimum + K; // Traverse the array A[] for (int i = 0; i < N; i++) { // If required criteria is not satisfied if (abs(arr[i] - ans) > K) { // Update ans ans = -1; break; } } // Print the result cout << ans;}// Driver Codeint main(){ int arr[] = { 1, 2, 5 }; int K = 2; int N = sizeof(arr) / sizeof(arr[0]); maximumNumber(arr, N, K); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find maximum value// of X such that |A[i] - X| ? Kstatic void maximumNumber(int arr[], int N, int K){ // Stores the smallest array element int minimum = Arrays.stream(arr).min().getAsInt(); // Store the possible value of X int ans = minimum + K; // Traverse the array A[] for(int i = 0; i < N; i++) { // If required criteria is not satisfied if (Math.abs(arr[i] - ans) > K) { // Update ans ans = -1; break; } } // Print the result System.out.print(ans);}// Driver Codepublic static void main(String args[]){ int arr[] = { 1, 2, 5 }; int K = 2; int N = arr.length; maximumNumber(arr, N, K);}}// This code is contributed by sanjoy_62 |
Python3
# Python3 program for the above approach# Function to find maximum value# of X such that |A[i] - X| ? Kdef maximumNumber(arr, N, K): # Stores the smallest array element minimum = min(arr) # Store the possible value of X ans = minimum + K # Traverse the array A[] for i in range(N): # If required criteria is not satisfied if (abs(arr[i] - ans) > K): # Update ans ans = -1 break # Print the result print(ans)# Driver Codeif __name__ == '__main__': arr = [1, 2, 5] K = 2 N = len(arr) maximumNumber(arr, N, K)# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find maximum value// of X such that |A[i] - X| ? Kstatic void maximumNumber(int []arr, int N, int K){ // Stores the smallest array element int mn = 100000000; for(int i = 0; i < N; i++) { if (arr[i] < mn) mn = arr[i]; } // Store the possible value of X int ans = mn + K; // Traverse the array A[] for(int i = 0; i < N; i++) { // If required criteria is not satisfied if (Math.Abs(arr[i] - ans) > K) { // Update ans ans = -1; break; } } // Print the result Console.Write(ans);}// Driver Codepublic static void Main(){ int []arr = { 1, 2, 5 }; int K = 2; int N = arr.Length; maximumNumber(arr, N, K);}}// This code is contributed by ipg2016107 |
Javascript
<script> // Javascript program for // the above approach // Function to find maximum value // of X such that |A[i] - X| ? K function maximumNumber(arr, N, K) { // Stores the smallest // array element let minimum = Math.min(...arr) // Store the possible value of X let ans = minimum + K; // Traverse the array A[] for (let i = 0; i < N; i++) { // If required criteria is // not satisfied if (Math.abs(arr[i] - ans) > K) { // Update ans ans = -1; break; } } // Print the result document.write(ans) } // Driver Code let arr = [1, 2, 5] let K = 2 let N = arr.length maximumNumber(arr, N, K); // This code is contributed by Hritik </script> |
Output
3
Time Complexity: O(N)
Auxiliary Space: O(1)
Approach 2: Binary Search:
Another approach to solve this problem is to use binary search. We can find the range of possible values of X using the smallest and largest elements of the array. Then, we can perform a binary search in this range to find the maximum value of X that satisfies the given condition.
Here is the implementation of this approach:
C++
#include <bits/stdc++.h>using namespace std;// Function to find maximum value// of X such that |A[i] - X| ? Kint maximumNumber(int arr[], int N, int K){ // Stores the smallest and largest array elements int minimum = *min_element(arr, arr + N); int maximum = *max_element(arr, arr + N); // Store the range of possible values of X int low = minimum + K; int high = maximum - K; // Perform binary search to find // maximum value of X that satisfies // the given condition while (low <= high) { int mid = low + (high - low) / 2; bool possible = true; for (int i = 0; i < N; i++) { if (abs(arr[i] - mid) > K) { possible = false; break; } } if (possible) { low = mid + 1; } else { high = mid - 1; } } // Return the maximum value of X return high;}// Driver Codeint main(){ int arr[] = {1, 2, 5}; int K = 2; int N = sizeof(arr) / sizeof(arr[0]); int ans = maximumNumber(arr, N, K); if (ans == -1) { cout << "No such X exists"; } else { cout << ans; } return 0;} |
Java
import java.util.*;public class Main {// Function to find maximum value// of X such that |A[i] - X| ? Kpublic static int maximumNumber(int arr[], int N, int K) { // Stores the smallest and largest array elements int minimum = Arrays.stream(arr).min().getAsInt(); int maximum = Arrays.stream(arr).max().getAsInt(); // Store the range of possible values of X int low = minimum + K; int high = maximum - K; // Perform binary search to find // maximum value of X that satisfies // the given condition while (low <= high) { int mid = low + (high - low) / 2; boolean possible = true; for (int i = 0; i < N; i++) { if (Math.abs(arr[i] - mid) > K) { possible = false; break; } } if (possible) { low = mid + 1; } else { high = mid - 1; } } // Return the maximum value of X return high;}// Driver Codepublic static void main(String[] args) { int arr[] = {1, 2, 5}; int K = 2; int N = arr.length; int ans = maximumNumber(arr, N, K); if (ans == -1) { System.out.println("No such X exists"); } else { System.out.println(ans); }}} |
Python3
# Function to find maximum value# of X such that |A[i] - X| ? Kdef maximumNumber(arr, N, K): # Stores the smallest and largest array elements minimum = min(arr) maximum = max(arr) # Store the range of possible values of X low = minimum + K high = maximum - K # Perform binary search to find # maximum value of X that satisfies # the given condition while low <= high: mid = low + (high - low) // 2 possible = True for i in range(N): if abs(arr[i] - mid) > K: possible = False break if possible: low = mid + 1 else: high = mid - 1 # Return the maximum value of X return high# Driver Codearr = [1, 2, 5]K = 2N = len(arr)ans = maximumNumber(arr, N, K)if ans == -1: print("No such X exists")else: print(ans) |
C#
using System;using System.Linq;class MainClass { public static int maximumNumber(int[] arr, int N, int K) { // Stores the smallest and largest array elements int minimum = arr.Min(); int maximum = arr.Max(); // Store the range of possible values of X int low = minimum + K; int high = maximum - K; // Perform binary search to find // maximum value of X that satisfies // the given condition while (low <= high) { int mid = low + (high - low) / 2; bool possible = true; for (int i = 0; i < N; i++) { if (Math.Abs(arr[i] - mid) > K) { possible = false; break; } } if (possible) { low = mid + 1; } else { high = mid - 1; } } // Return the maximum value of X return high; } public static void Main() { int[] arr = { 1, 2, 5 }; int K = 2; int N = arr.Length; int ans = maximumNumber(arr, N, K); if (ans == -1) { Console.WriteLine("No such X exists"); } else { Console.WriteLine(ans); } }}// This code is contributed by sarojmcy2e |
Javascript
// Function to find maximum value// of X such that |A[i] - X| ? Kfunction maximumNumber(arr, K) { const N = arr.length; // Stores the smallest and largest array elements const minimum = Math.min(...arr); const maximum = Math.max(...arr); // Store the range of possible values of X let low = minimum + K; let high = maximum - K; // Perform binary search to find // maximum value of X that satisfies // the given condition while (low <= high) { const mid = low + Math.floor((high - low) / 2); let possible = true; for (let i = 0; i < N; i++) { if (Math.abs(arr[i] - mid) > K) { possible = false; break; } } if (possible) { low = mid + 1; } else { high = mid - 1; } } // Return the maximum value of X return high;}// Driver Codeconst arr = [1, 2, 5];const K = 2;const ans = maximumNumber(arr, K);if (ans == -1) { console.log("No such X exists");} else { console.log(ans);}// This code is contributed by Sundaram |
Output
3
Time Complexity: O(N Log N)
Auxiliary Space: O(1)
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