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+1
Input : 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 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 code int 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; } |
chevron_right
filter_none
Java
// Java implementation to count ways // to sum up to a given value N 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(String[] args) { int N = 7; System.out.println("Total number of ways = " + countWays(N)); } } |
chevron_right
filter_none
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 Code arr = [1, 5, 6] m = len(arr) N = 7print("Total number of ways = ", countWays(arr, m, N)) # This code is contributed by Anant Agarwal. |
chevron_right
filter_none
C#
// C# implementation to count ways // to sum up to a given value N using 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. |
chevron_right
filter_none
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 ?> |
chevron_right
filter_none
Output:
Total number of ways = 6
Time Complexity: O(N*m)
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Reorder an Array according to given indices with repetition allowed
- Unbounded Knapsack (Repetition of items allowed)
- Number of ways to choose elements from the array such that their average is K
- Number of ways an array can be filled with 0s and 1s such that no consecutive elements are 1
- Number of ways to form an array with distinct adjacent elements
- Number of ways to choose an integer such that there are exactly K elements greater than it in the given array
- Minimum cost to reach end of array array when a maximum jump of K index is allowed
- Maximum sum from three arrays such that picking elements consecutively from same is not allowed
- Move all negative elements to end in order with extra space allowed
- Find maximum value of Sum( i*arr[i]) with only rotations on given array allowed
- Maximum array sum with prefix and suffix multiplications with -1 allowed
- Check if array contains contiguous integers with duplicates allowed
- Maximum sum of pairwise product in an array with negative allowed
- Find a Fixed Point in an array with duplicates allowed
- Check if array contains contiguous integers with duplicates allowed
- Maximum sum of values in a given range of an Array for Q queries when shuffling is allowed
- Check if it is possible to sort an array with conditional swapping of adjacent allowed
- Find a Fixed Point (Value equal to index) in a given array | Duplicates Allowed
- Count of Array elements greater than or equal to twice the Median of K trailing Array elements
- Number of ways to change the Array such that largest element is LCM of array
