Given two arrays div and rem which hold the values for divisors and remainders, the task is to find the number which after being successively divided by the elements of the div array leaves remainder which are in the rem array.
Note: The quotient from the first division will be divided by the second element and then the resultant quotient will be divided by the third (matching the remainders given) and so on.
Examples:
Input: div[] = {3, 5, 7}, rem[] = {1, 3, 5}
Output: 85
85 on division with 3 leaves remainder 1 with quotient 28
28 on division with 5 leaves remainder 3 with quotient 5
5 on division with 7 leaves remainder 5Input: div[] = {7, 9}, rem[] = {2, 2}
Output: 16
Approach:
- Store the value of the last remainder in a variable say num.
- Traverse the array backwards from n-2 to 0 and update the num as num = num * div[i] + rem[i]
- Print num in the end.
Below is the implementation of the above approach:
C++
// C++ program to implement above approach#include <bits/stdc++.h>using namespace std;// Function to find the numberint findNum(int div[], int rem[], int N){ int num = rem[N - 1]; for (int i = N - 2; i >= 0; i--) { num = num * div[i] + rem[i]; } return num;}// Driver Codeint main(){ int div[] = { 8, 3 }; int rem[] = { 2, 2 }; int N = sizeof(div) / sizeof(div[0]); cout << findNum(div, rem, N); return 0;} |
Java
// Java implementation of the above approachpublic class GFG{ // Function to find the number static int findNum(int div[], int rem[], int N) { int num = rem[N - 1]; for (int i = N - 2; i >= 0; i--) { num = num * div[i] + rem[i]; } return num; } // Driver Code public static void main(String []args){ int div[] = { 8, 3 }; int rem[] = { 2, 2 }; int N = div.length; System.out.println(findNum(div, rem, N)); } // This code is contributed by ANKITRAI1} |
Python3
# Python 3 program to implement# above approach# Function to find the numberdef findNum(div, rem, N): num = rem[N - 1] i = N - 2 while(i >= 0): num = num * div[i] + rem[i] i -= 1 return num# Driver Codeif __name__ == '__main__': div = [8, 3] rem = [2, 2] N = len(div) print(findNum(div, rem, N))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the // above approach using System;class GFG{// Function to find the number static int findNum(int []div, int []rem, int N) { int num = rem[N - 1]; for (int i = N - 2; i >= 0; i--) { num = num * div[i] + rem[i]; } return num; } // Driver Code static public void Main (){ int []div = { 8, 3 }; int []rem = { 2, 2 }; int N = div.Length; Console.WriteLine(findNum(div, rem, N)); } } // This code is contributed by ajit |
PHP
<?php// PHP program to implement// above approach// Function to find the numberfunction findNum($div, $rem, $N){ $num = $rem[$N - 1]; for ($i = $N - 2; $i >= 0; $i--) { $num = $num * $div[$i] + $rem[$i]; } return $num;}// Driver Code$div = array( 8, 3 );$rem = array(2, 2 );$N = sizeof($div);echo findNum($div, $rem, $N);// This code is contributed by ajit?> |
Javascript
<script>// Javascript program to implement above approach// Function to find the numberfunction findNum(div, rem, N){ var num = rem[N - 1]; for (var i = N - 2; i >= 0; i--) { num = num * div[i] + rem[i]; } return num;}// Driver Codevar div = [ 8, 3 ];var rem = [ 2, 2 ];var N = div.length;document.write( findNum(div, rem, N));</script> |
18
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
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