Given an array arr[] and two integers N and X, the task is to find if it is possible to create an array using N distinct elements from arr[] such that the mean of the newly formed array is X.
Examples:
Input: N = 5, X = 8, arr[] = {1, 10, 3, 2, 6, 7, 4, 5}
Output: YES
Explanation: Many arrays using 5 distinct elements from the array are possible like {10, 6, 7, 4, 5, 10, 10, 10, 10}Input: N = 3, X = 4, arr[] = {9, 7, 5}
Output: NO
Explanation: There is no possible array by a given finite set. So the mean of the array becomes exact to X.Therefore, the answer is NO.
Approach: Implement the idea below to solve the problem:
It is always possible to make mean X from a given arr[] of the integer if it lies in between the minimum and maximum integer in an arr[]. Formally, if ( X >= MinValue && X <= MaxValue ) then, the answer will be YES only for those cases otherwise NO.
Steps were taken to solve the problem:
- Create an integer variable min and store the minimum value in this variable by iterating on arr[].
- Create an integer variable max and store the maximum value in this variable by iterating on arr[].
- if X lies in the range of min and max value of arr[], Then print YES else print NO.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to check if it is possible// to create the array satisfying the conditionsstring isPossible(int N, int X, int arr[]){ // Min variable to store minimum // value from input arr[] int sz = sizeof(arr) / sizeof(arr[0]); int min = INT_MAX; // Max variable to store maximum // value from input arr[] int max = INT_MIN; // Loop for iterating on arr[] for (int i = 0; i < sz; i++) { // Updating value of min variable min = arr[i] < min ? arr[i] : min; // Updating value of max variable max = arr[i] > max ? arr[i] : max; } // Printing answer as YES/NO by // comparing mean with min and max // values return X >= min && X <= max ? "YES" : "NO";}// Driver Codeint main(){ int N = 5, X = 8; int arr[] = { 1, 10, 3, 2, 6, 7, 4, 5 }; // Function call cout << isPossible(N, X, arr); return 0;}// This code is contributed by rohit768. |
Java
// Java code to implement the approachimport java.io.*;// Driver Classclass GFG { // Driver code public static void main(String[] args) { int N = 5, X = 8; int[] arr = { 1, 10, 3, 2, 6, 7, 4, 5 }; // Function call System.out.println(isPossible(N, X, arr)); } // Function to check if it is possible // to create the array satisfying the conditions static String isPossible(int N, int X, int arr[]) { // Min variable to store minimum // value from input arr[] int min = Integer.MAX_VALUE; // Max variable to store maximum // value from input arr[] int max = Integer.MIN_VALUE; // Loop for iterating on arr[] for (int i = 0; i < arr.length; i++) { // Updating value of min variable min = arr[i] < min ? arr[i] : min; // Updating value of max variable max = arr[i] > max ? arr[i] : max; } // Printing answer as YES/NO by // comparing mean with min and max // values return X >= min && X <= max ? "YES" : "NO"; }} |
Python3
# python code to implement the approach# Function to check if it is possible# to create the array satisfying the conditionsdef isPossible(N, X, arr): # Min variable to store minimum # value from input arr[] sz = len(arr) mini = 9223372036854775807 # Max variable to store maximum # value from input arr[] maxi = -9223372036854775808 # Loop for iterating on arr[] for i in range(0, sz): # Updating value of min variable if(arr[i] < mini): mini = arr[i] # Updating value of max variable if(arr[i] > maxi): maxi = arr[i] # Print answer as YES/NO by # comparing mean with min and max # values if(X >= mini and X <= maxi): return "YES" else: return "NO"# Driver CodeN = 5X = 8arr = [1, 10, 3, 2, 6, 7, 4, 5]# Function callprint(isPossible(N, X, arr))# This code is contributed by ksam24000 |
C#
// C# code to implement the approachusing System;// Driver Classclass GFG { // Driver code public static void Main() { int N = 5, X = 8; int[] arr = { 1, 10, 3, 2, 6, 7, 4, 5 }; // Function call Console.WriteLine(isPossible(N, X, arr)); } // Function to check if it is possible // to create the array satisfying the conditions static string isPossible(int N, int X, int[] arr) { // Min variable to store minimum // value from input arr[] int min = Int32.MaxValue; // Max variable to store maximum // value from input arr[] int max = Int32.MinValue; // Loop for iterating on arr[] for (int i = 0; i < arr.Length; i++) { // Updating value of min variable min = arr[i] < min ? arr[i] : min; // Updating value of max variable max = arr[i] > max ? arr[i] : max; } // Printing answer as YES/NO by // comparing mean with min and max // values return X >= min && X <= max ? "YES" : "NO"; }}// This code is contributed by Samim Hossain Mondal. |
Javascript
// JS code to implement the approach// Function to check if it is possible// to create the array satisfying the conditionsfunction isPossible(N,X,arr){ // Min variable to store minimum // value from input arr[] let sz = arr.length; let mn = Number.MAX_VALUE; // Max variable to store maximum // value from input arr[] let mx = Number.MIN_VALUE; // Loop for iterating on arr[] for (let i = 0; i < sz; i++) { // Updating value of min variable mn = arr[i] < mn ? arr[i] : mn; // Updating value of max variable mx = arr[i] > mx ? arr[i] : mx; } // Printing answer as YES/NO by // comparing mean with min and max // values return X >= mn && X <= mx ? "YES" : "NO";}// Driver Code let N = 5, X = 8; let arr = [ 1, 10, 3, 2, 6, 7, 4, 5 ]; // Function call console.log(isPossible(N, X, arr)); // This code is contributed by ksam24000. |
Output
YES
Time Complexity: O(N)
Auxiliary Space: O(1)
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