Given an array of amounts and time_period that represents the amount of money N persons invest and the time period for which they had invested. The task is to calculate the Profit ratio at the End.
Examples:
Input: n = 2,
Amount1 = 7000, Time1 = 12 months
Amount2 = 6000, Time2 = 6 months
Output: 7 : 3
Input: n = 3,
Amount1 = 5000, Time1 = 6 months
Amount2 = 6000, Time2 = 6 months
Amount3 = 1000, Time3 = 12 months
Output: 5 : 6: 2
Formula:
1st person share: (Amount of money invested by 1st) * (Time Period of 1st)
2nd Person share: (Amount of money invested by 2nd) * (Time Period of 2nd)
3rd Person share: (Amount of money invested by 3rd) * (Time Period of 3rd) and so on…
Ratio: 1st person share: 2nd person share : 3rd Person Share
Below is the required implementation:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Calculating GCD of an array.int find_Gcd(int crr[], int n){ int i; int result = crr[0]; for (i = 1; i < n; i++) result = __gcd(crr[i], result); return result;}// Function to calculate the Sharevoid profitRatio(int amountArr[], int timeTrr[], int n){ int i, crr[n]; for (i = 0; i < n; i++) crr[i] = amountArr[i] * timeTrr[i]; int Share = find_Gcd(crr, n); for (i = 0; i < n - 1; i++) cout << crr[i] / Share << " : "; cout << crr[i] / Share;}// Driver Codeint main(){ int amountArr[] = { 5000, 6000, 1000 }; int timeTrr[] = { 6, 6, 12 }; int n = sizeof(amountArr) / sizeof(amountArr[0]); profitRatio(amountArr, timeTrr, n); return 0;} |
Java
// Java implementation of // above approachimport java.io.*;class GFG {// Recursive function to // return gcd of a and b static int __gcd(int a, int b) { // Everything divides 0 if (a == 0 || b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } // Calculating GCD of an array.static int find_Gcd(int crr[], int n){ int i; int result = crr[0]; for (i = 1; i < n; i++) result = __gcd(crr[i], result); return result;}// Function to calculate the Sharestatic void profitRatio(int amountArr[], int timeTrr[], int n){ int i; int crr[] = new int[n] ; for (i = 0; i < n; i++) crr[i] = amountArr[i] * timeTrr[i]; int Share = find_Gcd(crr, n); for (i = 0; i < n - 1; i++) System.out.print(crr[i] / Share + " : "); System.out.print(crr[i] / Share);}// Driver Codepublic static void main (String[] args) { int amountArr[] = {5000, 6000, 1000}; int timeTrr[] = {6, 6, 12}; int n = amountArr.length; profitRatio(amountArr, timeTrr, n);}}// This code is contributed // by inder_verma. |
Python3
# Python3 implementation of above approach# Recursive function to # return gcd of a and b def __gcd(a, b): # Everything divides 0 if(a == 0 or b == 0): return 0; # base case if(a == b): return a; # a is greater if(a > b): return __gcd(a - b, b); return __gcd(a, b - a); # Calculating GCD of an array.def find_Gcd(crr, n): result = crr[0]; for i in range(1, n): result = __gcd(crr[i], result); return result;# Function to calculate the Sharedef profitRatio(amountArr, timeTrr, n): i = 0; crr = [0] * n; for i in range(n): crr[i] = amountArr[i] * timeTrr[i]; Share = find_Gcd(crr, n); for i in range(n - 1): print(int(crr[i] / Share), end = " : "); print(int(crr[i + 1] / Share));# Driver CodeamountArr = [5000, 6000, 1000];timeTrr = [6, 6, 12];n = len(amountArr); profitRatio(amountArr, timeTrr, n);# This code is contributed # by mits |
C#
// C# implementation of // above approachusing System;class GFG {// Recursive function to // return gcd of a and b static int __gcd(int a, int b) { // Everything divides 0 if (a == 0 || b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } // Calculating GCD of an array.static int find_Gcd(int []crr, int n){ int i; int result = crr[0]; for (i = 1; i < n; i++) result = __gcd(crr[i], result); return result;}// Function to calculate the Sharestatic void profitRatio(int []amountArr, int []timeTrr, int n){ int i; int []crr = new int[n] ; for (i = 0; i < n; i++) crr[i] = amountArr[i] * timeTrr[i]; int Share = find_Gcd(crr, n); for (i = 0; i < n - 1; i++) Console.Write(crr[i] / Share + " : "); Console.Write(crr[i] / Share);}// Driver Codepublic static void Main () { int []amountArr = {5000, 6000, 1000}; int []timeTrr = {6, 6, 12}; int n = amountArr.Length; profitRatio(amountArr, timeTrr, n);}}// This code is contributed // by inder_verma. |
PHP
<?php// PHP implementation of // above approach// Recursive function to // return gcd of a and b function __gcd($a, $b) { // Everything divides 0 if ($a == 0 or $b == 0) return 0; // base case if ($a == $b) return $a; // a is greater if ($a > $b) return __gcd($a - $b, $b); return __gcd($a, $b - $a); } // Calculating GCD of an array.function find_Gcd($crr, $n){ $i; $result = $crr[0]; for ($i = 1; $i < $n; $i++) $result = __gcd($crr[$i], $result); return $result;}// Function to calculate the Sharefunction profitRatio($amountArr, $timeTrr, $n){ $i; $crr = array(); for ($i = 0; $i < $n; $i++) $crr[$i] = $amountArr[$i] * $timeTrr[$i]; $Share = find_Gcd($crr, $n); for ($i = 0; $i < $n - 1; $i++) echo $crr[$i] / $Share , " : "; echo $crr[$i] / $Share;}// Driver Code$amountArr = array(5000, 6000, 1000);$timeTrr = array(6, 6, 12);$n = count($amountArr); profitRatio($amountArr, $timeTrr, $n);// This code is contributed // by inder_verma?> |
Javascript
<script>// javascript implementation of // above approach // Recursive function to // return gcd of a and b function __gcd(a , b) { // Everything divides 0 if (a == 0 || b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } // Calculating GCD of an array. function find_Gcd(crr , n) { var i; var result = crr[0]; for (i = 1; i < n; i++) result = __gcd(crr[i], result); return result; } // Function to calculate the Share function profitRatio(amountArr , timeTrr , n) { var i; var crr = Array(n).fill(0); for (i = 0; i < n; i++) crr[i] = amountArr[i] * timeTrr[i]; var Share = find_Gcd(crr, n); for (i = 0; i < n - 1; i++) document.write(crr[i] / Share + " : "); document.write(crr[i] / Share); } // Driver Code var amountArr = [ 5000, 6000, 1000 ]; var timeTrr = [ 6, 6, 12 ]; var n = amountArr.length; profitRatio(amountArr, timeTrr, n);// This code contributed by aashish1995</script> |
Output:
5 : 6 : 2
Time Complexity: O(nlogm) where m is the max value of the array crr.
Auxiliary Space: O(n)
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