Given an array arr[], the task is to find the maximum ratio pair in the array.
Examples:
Input: arr[] = { 15, 10, 3 }
Output: 5
Explanation:
Maximum ratio pair will be –
Input: arr[] = { 15, 10, 3, 2 }
Output: 7.5
Explanation:
Maximum ratio pair will be –
Approach: The idea is to iterate over every possible pair of the array using two nested loops and find the maximum ratio pair possible. For any pair, the maximum ratio can be obtained using 
Below is the implementation of the above approach:
C++
// C++ implementation to find// the maximum pair in the array#include <bits/stdc++.h>using namespace std;// Function to find the maximum pair// possible for the arrayfloat computeMaxValue(float arr[], int n){ float ans = 0; // Loop to iterate over every // possible pair in the array for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { // Check pair (x, y) as well as // (y, x) for maximum value float val = max(arr[i] / arr[j], arr[j] / arr[i]); // Update the answer ans = max(ans, val); } } return ans;}// Driver Codeint main(){ float arr[] = { 15, 10, 3, 2 }; int n = sizeof(arr) / sizeof(arr[0]); cout << computeMaxValue(arr, n); return 0;} |
Java
// Java implementation to find // the maximum pair in the array import java.io.*; import java.util.*; class GFG { // Function to find the maximum pair // possible for the array static float computeMaxValue(float arr[], int n) { float ans = 0; // Loop to iterate over every // possible pair in the array for(int i = 0; i < n - 1; i++) { for(int j = i + 1; j < n; j++) { // Check pair (x, y) as well as // (y, x) for maximum value float val = Math.max(arr[i] / arr[j], arr[j] / arr[i]); // Update the answer ans = Math.max(ans, val); } } return ans; }// Driver code public static void main(String[] args) { float arr[] = { 15, 10, 3, 2 }; int N = arr.length; System.out.println(computeMaxValue(arr, N)); } } // This code is contributed by coder001 |
Python3
# Python3 implementation to find# the maximum pair in the array# Function to find the maximum pair# possible for the arraydef computeMaxValue(arr, n): ans = 0 # Loop to iterate over every # possible pair in the array for i in range(n - 1): for j in range(i + 1, n): # Check pair (x, y) as well as # (y, x) for maximum value val = max(arr[i] / arr[j], arr[j] / arr[i]) # Update the answer ans = max(ans, val) return ans# Driver Codeif __name__ == "__main__": arr = [ 15, 10, 3, 2 ] n = len(arr) print(computeMaxValue(arr, n)) # This code is contributed by chitranayal |
C#
// C# implementation to find // the maximum pair in the array using System;class GFG { // Function to find the maximum pair // possible for the array static float computeMaxValue(float []arr, int n) { float ans = 0; // Loop to iterate over every // possible pair in the array for(int i = 0; i < n - 1; i++) { for(int j = i + 1; j < n; j++) { // Check pair (x, y) as well as // (y, x) for maximum value float val = Math.Max(arr[i] / arr[j], arr[j] / arr[i]); // Update the answer ans = Math.Max(ans, val); } } return ans; }// Driver code public static void Main(String[] args) { float []arr = { 15, 10, 3, 2 }; int N = arr.Length; Console.WriteLine(computeMaxValue(arr, N)); } }// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation to find// the maximum pair in the array// Function to find the maximum pair// possible for the arrayfunction computeMaxValue(arr, n){ var ans = 0; // Loop to iterate over every // possible pair in the array for (var i = 0; i < n - 1; i++) { for (var j = i + 1; j < n; j++) { // Check pair (x, y) as well as // (y, x) for maximum value var val = Math.max(arr[i] / arr[j], arr[j] / arr[i]); // Update the answer ans = Math.max(ans, val); } } return ans;}// Driver Codevar arr = [ 15, 10, 3, 2 ];var n = arr.length;document.write( computeMaxValue(arr, n));</script> |
Output
7.5
Time Complexity: O(N2)
Auxiliary Space: O(1)
Approach-2 For a[i]>0
As the maximum ratio can be obtained when the denominator is minimum.
- So store the index of the minimum element of the array.
- Loop through the array and divide each integer with the minimum number.
- Maintain a running maximum.
Below is the implementation of the above approach:
C++
// C++ implementation to find// the maximum pair in the array#include <bits/stdc++.h>using namespace std;// Function to find the index of minimum number of arrayint minNum(float a[], int n){ int indOfMin = 0; // initializing with 0 for (int i = 0; i < n; i++) { if (a[i] < a[indOfMin]) indOfMin = i; } return indOfMin;}// Function to find the maximum pair// possible for the arrayfloat computeMaxValue(float a[], int n){ int minIndex = minNum(a, n); float ans = INT_MIN; for (int i = 0; i < n; i++) { if (i != minIndex) { float temp = float(a[i]) / float(a[minIndex]); ans = max(ans, temp); } } return ans;}// Driver Codeint main(){ float arr[] = { 15, 10, 3, 2 }; int n = sizeof(arr) / sizeof(arr[0]); cout << computeMaxValue(arr, n); return 0;}// This code is contributed by Palak Gupta |
Java
// Java implementation to find// the maximum pair in the arrayimport java.io.*;import java.util.*;class GFG { // Function to find the index of minimum number of array static int minNum(float a[], int n) { int indOfMin = 0; // initializing with 0 for (int i = 0; i < n; i++) { if (a[i] < a[indOfMin]) indOfMin = i; } return indOfMin; } // Function to find the maximum pair // possible for the array static float computeMaxValue(float a[], int n) { int minIndex = minNum(a, n); float ans = Integer.MIN_VALUE; for (int i = 0; i < n; i++) { if (i != minIndex) { float temp = (float)a[i] / (float)a[minIndex]; if (temp > ans) ans = temp; } } return ans; } // Driver code public static void main(String[] args) { float arr[] = { 15, 10, 3, 2 }; int N = arr.length; System.out.println(computeMaxValue(arr, N)); }}// This code is contributed by Palak Gupta |
Python3
# Python implementation to find# the maximum pair in the array# Function to find the index of minimum number of arraydef minNum(a, n): indOfMin = 0 # initializing with 0 for i in range(0, n): if (a[i] < a[indOfMin]): indOfMin = i return indOfMin# Function to find the maximum pair# possible for the arraydef computeMaxValue(a, n): minIndex = minNum(a, n) ans = float('-inf') for i in range(0, n): if (i != minIndex): temp = float(a[i]) / float(a[minIndex]) ans = max(ans, temp) return ans# Driver Codearr = [15, 10, 3, 2]n = len(arr)print(computeMaxValue(arr, n))# This code is contributed by ninja_hattori. |
C#
using System;class GFG{ // Function to find the index of minimum number of array static int minNum(float []a, int n) { int indOfMin = 0; // initializing with 0 for (int i = 0; i < n; i++) { if (a[i] < a[indOfMin]) indOfMin = i; } return indOfMin; } // Function to find the maximum pair // possible for the array static float computeMaxValue(float []a, int n) { int minIndex = minNum(a, n); float ans = Int32.MinValue; for (int i = 0; i < n; i++) { if (i != minIndex) { float temp = (float)a[i] / (float)a[minIndex]; if (temp > ans) ans = temp; } } return ans; } // Driver code public static void Main(String[] args) { float []arr = { 15, 10, 3, 2 }; int N = arr.Length; Console.WriteLine(computeMaxValue(arr, N)); }}// This code is contributed by Kritima Gupta |
Javascript
<script>// Javascript implementation to find// the maximum pair in the array// Function to find the index of minimum number of arrayfunction minNum(a, n){ var indOfMin = 0; // initializing with 0 for (var i = 0; i < n; i++) { if (a[i] < a[indOfMin]) indOfMin = i; } return indOfMin;}// Function to find the maximum pair// possible for the arrayfunction computeMaxValue( a, n){ var minIndex = minNum(a, n); var ans = Number.MIN_VALUE; for (var i = 0; i < n; i++) { if (i != minIndex) { var temp = Number(a[i]) / Number(a[minIndex]); ans = Math.max(ans, temp); } } return ans;}// Driver Codevar arr = [15, 10, 3, 2 ];var n = arr.length;document.write(computeMaxValue(arr, n));// This code is contributed by ninja_hattori.</script> |
Output
7.5
Time Complexity: O(N)
Auxiliary Space: O(1)
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