Given a set of m distinct positive integers and a value ‘N’. The problem is to count the total number of ways we can form ‘N’ by doing sum of the array elements. Repetitions and different arrangements are allowed.
Examples :
Input: arr = {1, 5, 6}, N = 7
Output: 6
Explanation: The different ways are:
1+1+1+1+1+1+1
1+1+5
1+5+1
5+1+1
1+6
6+1Input: arr = {12, 3, 1, 9}, N = 14
Output: 150
Approach: The approach is based on the concept of dynamic programming.
countWays(arr, m, N)
Declare and initialize count[N + 1] = {0}
count[0] = 1
for i = 1 to N
for j = 0 to m - 1
if i >= arr[j]
count[i] += count[i - arr[j]]
return count[N]
Below is the implementation of the above approach.
C++
// C++ implementation to count ways // to sum up to a given value N#include <bits/stdc++.h>using namespace std;// function to count the total // number of ways to sum up to 'N'int countWays(int arr[], int m, int N){ int count[N + 1]; memset(count, 0, sizeof(count)); // base case count[0] = 1; // count ways for all values up // to 'N' and store the result for (int i = 1; i <= N; i++) for (int j = 0; j < m; j++) // if i >= arr[j] then // accumulate count for value 'i' as // ways to form value 'i-arr[j]' if (i >= arr[j]) count[i] += count[i - arr[j]]; // required number of ways return count[N]; }// Driver codeint main(){ int arr[] = {1, 5, 6}; int m = sizeof(arr) / sizeof(arr[0]); int N = 7; cout << "Total number of ways = " << countWays(arr, m, N); return 0;} |
Java
// Java implementation to count ways // to sum up to a given value Nclass Gfg{ static int arr[] = {1, 5, 6}; // method to count the total number // of ways to sum up to 'N' static int countWays(int N) { int count[] = new int[N + 1]; // base case count[0] = 1; // count ways for all values up // to 'N' and store the result for (int i = 1; i <= N; i++) for (int j = 0; j < arr.length; j++) // if i >= arr[j] then // accumulate count for value 'i' as // ways to form value 'i-arr[j]' if (i >= arr[j]) count[i] += count[i - arr[j]]; // required number of ways return count[N]; } // Driver code public static void main(String[] args) { int N = 7; System.out.println("Total number of ways = " + countWays(N)); }} |
Python3
# Python3 implementation to count # ways to sum up to a given value N# Function to count the total # number of ways to sum up to 'N'def countWays(arr, m, N): count = [0 for i in range(N + 1)] # base case count[0] = 1 # Count ways for all values up # to 'N' and store the result for i in range(1, N + 1): for j in range(m): # if i >= arr[j] then # accumulate count for value 'i' as # ways to form value 'i-arr[j]' if (i >= arr[j]): count[i] += count[i - arr[j]] # required number of ways return count[N] # Driver Codearr = [1, 5, 6]m = len(arr)N = 7print("Total number of ways = ", countWays(arr, m, N)) # This code is contributed by Anant Agarwal. |
C#
// C# implementation to count ways // to sum up to a given value Nusing System;class Gfg{ static int []arr = {1, 5, 6}; // method to count the total number // of ways to sum up to 'N' static int countWays(int N) { int []count = new int[N+1]; // base case count[0] = 1; // count ways for all values up // to 'N' and store the result for (int i = 1; i <= N; i++) for (int j = 0; j < arr.Length; j++) // if i >= arr[j] then // accumulate count for value 'i' as // ways to form value 'i-arr[j]' if (i >= arr[j]) count[i] += count[i - arr[j]]; // required number of ways return count[N]; } // Driver code public static void Main() { int N = 7; Console.Write("Total number of ways = " + countWays(N)); }}//This code is contributed by nitin mittal. |
PHP
<?php// PHP implementation to count ways // to sum up to a given value N // function to count the total // number of ways to sum up to 'N' function countWays($arr, $m, $N) { $count = array_fill(0,$N + 1,0); // base case $count[0] = 1; // count ways for all values up // to 'N' and store the result for ($i = 1; $i <= $N; $i++) for ($j = 0; $j < $m; $j++) // if i >= arr[j] then // accumulate count for value 'i' as // ways to form value 'i-arr[j]' if ($i >= $arr[$j]) $count[$i] += $count[$i - $arr[$j]]; // required number of ways return $count[$N]; } // Driver code $arr = array(1, 5, 6); $m = count($arr);$N = 7;echo "Total number of ways = ",countWays($arr, $m, $N); // This code is contributed by Ryuga?> |
Javascript
<script> // JavaScript implementation to count ways // to sum up to a given value N let arr = [1, 5, 6]; // method to count the total number // of ways to sum up to 'N' function countWays(N) { let count = new Array(N+1); count.fill(0); // base case count[0] = 1; // count ways for all values up // to 'N' and store the result for (let i = 1; i <= N; i++) for (let j = 0; j < arr.length; j++) // if i >= arr[j] then // accumulate count for value 'i' as // ways to form value 'i-arr[j]' if (i >= arr[j]) count[i] += count[i - arr[j]]; // required number of ways return count[N]; } let N = 7; document.write("Total number of ways = " + countWays(N)); </script> |
Output
Total number of ways = 6
Time Complexity: O(N*m)
Auxiliary Space: O(N), as an additional array of size (N + 1) is required.
This article is contributed by Ayush Jauhari. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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