Given an array, the task is to find the minimum and maximum elements in the array which are divisible by a given number k.
Examples:
Input: arr[] = {12, 1235, 45, 67, 1}, k=5
Output: Minimum = 45, Maximum = 1235
Input: arr[] = {10, 1230, 45, 67, 1}, k=10
Output: Minimum = 10, Maximum = 1230
Approach:
- Take a min variable that stores the minimum element and initialize it with INT_MAX and compare it with every element of the array and update the next minimum element which is divisible by k.
- Take a max variable that stores the maximum element and initialize it with INT_MIN and compare it with every element of the array and update the next maximum element which is divisible by k.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum elementint getMin(int arr[], int n, int k){ int res = INT_MAX; for (int i = 0; i < n; i++) { if (arr[i] % k == 0) res = min(res, arr[i]); } return res;}// Function to find the maximum elementint getMax(int arr[], int n, int k){ int res = INT_MIN; for (int i = 1; i < n; i++) { if (arr[i] % k == 0) res = max(res, arr[i]); } return res;}// Driver codeint main(){ int arr[] = { 10, 1230, 45, 67, 1 }; int k = 10; int n = sizeof(arr) / sizeof(arr[0]); cout << "Minimum element of array which is divisible by k: " << getMin(arr, n, k) << "\n"; cout << "Maximum element of array which is divisible by k: " << getMax(arr, n, k); return 0;} |
C
// C implementation of the above approach#include <stdio.h>#include <limits.h>int max(int a, int b){ int max = a; if(max < b) max = b; return max;}int min(int a,int b){ int min = a; if(min > b) min = b; return min;}// Function to find the minimum elementint getMin(int arr[], int n, int k){ int res = INT_MAX; for (int i = 0; i < n; i++) { if (arr[i] % k == 0) res = min(res, arr[i]); } return res;}// Function to find the maximum elementint getMax(int arr[], int n, int k){ int res = INT_MIN; for (int i = 1; i < n; i++) { if (arr[i] % k == 0) res = max(res, arr[i]); } return res;}// Driver codeint main(){ int arr[] = { 10, 1230, 45, 67, 1 }; int k = 10; int n = sizeof(arr) / sizeof(arr[0]); printf("Minimum element of array which is divisible by k: %d\n",getMin(arr, n, k)); printf("Maximum element of array which is divisible by k: %d\n",getMax(arr, n, k)); return 0;}// This code is contributed by kothavvsaakash. |
Java
//Java implementation of the above approach class GFG {// Function to find the minimum element static int getMin(int arr[], int n, int k) { int res = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { if (arr[i] % k == 0) { res = Math.min(res, arr[i]); } } return res; }// Function to find the maximum element static int getMax(int arr[], int n, int k) { int res = Integer.MIN_VALUE; for (int i = 1; i < n; i++) { if (arr[i] % k == 0) { res = Math.max(res, arr[i]); } } return res; }// Driver code public static void main(String[] args) { int arr[] = {10, 1230, 45, 67, 1}; int k = 10; int n = arr.length; System.out.println("Minimum element of array which is divisible by k: " + getMin(arr, n, k)); System.out.println("Maximum element of array which is divisible by k: " + getMax(arr, n, k)); }}//This code contribute by Shikha Singh |
Python 3
# Python 3 implementation of the # above approachimport sys# Function to find the minimum elementdef getMin(arr, n, k): res = sys.maxsize for i in range(n): if (arr[i] % k == 0): res = min(res, arr[i]) return res# Function to find the maximum elementdef getMax(arr, n, k): res = 0 for i in range(1, n): if (arr[i] % k == 0): res = max(res, arr[i]) return res# Driver codeif __name__ == "__main__": arr = [ 10, 1230, 45, 67, 1 ] k = 10 n = len(arr) print("Minimum element of array which", "is divisible by k: ", getMin(arr, n, k)) print( "Maximum element of array which", "is divisible by k: ", getMax(arr, n, k))# This code is contributed # by ChitraNayal |
C#
// C# implementation of the above approach using System;class GFG{// Function to find the minimum element static int getMin(int []arr, int n, int k) { int res = int.MaxValue; for (int i = 0; i < n; i++) { if (arr[i] % k == 0) { res = Math.Min(res, arr[i]); } } return res;}// Function to find the maximum element static int getMax(int []arr, int n, int k) { int res = int.MinValue; for (int i = 1; i < n; i++) { if (arr[i] % k == 0) { res = Math.Max(res, arr[i]); } } return res;}// Driver code static public void Main (){ int []arr = {10, 1230, 45, 67, 1}; int k = 10; int n = arr.Length; Console.WriteLine("Minimum element of array " + "which is divisible by k: " + getMin(arr, n, k)); Console.WriteLine("Maximum element of array " + "which is divisible by k: " + getMax(arr, n, k));}}// This code is contributes by ajit |
PHP
<?php// PHP implementation of the above approach // Function to find the minimum element function getMin($arr, $n, $k) { $res = PHP_INT_MAX; for ($i = 0; $i < $n; $i++) { if ($arr[$i] % $k == 0) $res = min($res, $arr[$i]); } return $res; } // Function to find the maximum element function getMax($arr, $n, $k) { $res = PHP_INT_MIN; for ($i = 1; $i < $n; $i++) { if ($arr[$i] % $k == 0) $res = max($res, $arr[$i]); } return $res; } // Driver code $arr = array( 10, 1230, 45, 67, 1 ); $k = 10; $n = sizeof($arr); echo "Minimum element of array which is " . "divisible by k: ", getMin($arr, $n, $k) , "\n"; echo "Maximum element of array which is " . "divisible by k: ", getMax($arr, $n, $k); // This code is contributed by akt_mit?> |
Javascript
<script>// Javascript implementation of the above approach// Function to find the minimum elementfunction getMin(arr, n, k){ let res = Number.MAX_VALUE; for (let i = 0; i < n; i++) { if (arr[i] % k == 0) res = Math.min(res, arr[i]); } return res;}// Function to find the maximum elementfunction getMax(arr, n, k){ let res = Number.MIN_VALUE; for (let i = 1; i < n; i++) { if (arr[i] % k == 0) res = Math.max(res, arr[i]); } return res;}// Driver code let arr = [ 10, 1230, 45, 67, 1 ]; let k = 10; let n = arr.length; document.write("Minimum element of array which is divisible by k: " + getMin(arr, n, k) + "<br>"); document.write("Maximum element of array which is divisible by k: " + getMax(arr, n, k));</script> |
Output
Minimum element of array which is divisible by k: 10 Maximum element of array which is divisible by k: 1230
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
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!
