Given two natural number n and m. The task is to find the number of ways in which the numbers that are greater than or equal to m can be added to get the sum n.
Examples:
Input : n = 3, m = 1 Output : 3 Following are three different ways to get sum n such that each term is greater than or equal to m 1 + 1 + 1, 1 + 2, 3 Input : n = 2, m = 1 Output : 2 Two ways are 1 + 1 and 2
Method 1:
The idea is to use Dynamic Programming by define 2D matrix, say dp[][]. dp[i][j] define the number of ways to get sum i using the numbers greater than or equal to j. So dp[i][j] can be defined as:
If i < j, dp[i][j] = 0, because we cannot achieve smaller sum of i using numbers greater than or equal to j.
If i = j, dp[i][j] = 1, because there is only one way to show sum i using number i which is equal to j.
Else dp[i][j] = dp[i][j+1] + dp[i-j][j], because obtaining a sum i using numbers greater than or equal to j is equal to the sum of obtaining a sum of i using numbers greater than or equal to j+1 and obtaining the sum of i-j using numbers greater than or equal to j.
Below is the implementation of this approach:
C++
// CPP Program to find number of ways to// which numbers that are greater than// given number can be added to get sum.#include <bits/stdc++.h>#define MAX 100using namespace std;// Return number of ways to which numbers// that are greater than given number can// be added to get sum.int numberofways(int n, int m){ int dp[n+2][n+2]; memset(dp, 0, sizeof(dp)); dp[0][n + 1] = 1; // Filling the table. k is for numbers // greater than or equal that are allowed. for (int k = n; k >= m; k--) { // i is for sum for (int i = 0; i <= n; i++) { // initializing dp[i][k] to number // ways to get sum using numbers // greater than or equal k+1 dp[i][k] = dp[i][k + 1]; // if i > k if (i - k >= 0) dp[i][k] = (dp[i][k] + dp[i - k][k]); } } return dp[n][m];}// Driver Programint main(){ int n = 3, m = 1; cout << numberofways(n, m) << endl; return 0;} |
Java
// Java Program to find number of ways to// which numbers that are greater than// given number can be added to get sum.import java.io.*;class GFG { // Return number of ways to which numbers // that are greater than given number can // be added to get sum. static int numberofways(int n, int m) { int dp[][]=new int[n+2][n+2]; dp[0][n + 1] = 1; // Filling the table. k is for numbers // greater than or equal that are allowed. for (int k = n; k >= m; k--) { // i is for sum for (int i = 0; i <= n; i++) { // initializing dp[i][k] to number // ways to get sum using numbers // greater than or equal k+1 dp[i][k] = dp[i][k + 1]; // if i > k if (i - k >= 0) dp[i][k] = (dp[i][k] + dp[i - k][k]); } } return dp[n][m]; } // Driver Program public static void main(String args[]) { int n = 3, m = 1; System.out.println(numberofways(n, m)); }}/*This code is contributed by Nikita tiwari.*/ |
Python3
# Python3 Program to find number of ways to# which numbers that are greater than# given number can be added to get sum.MAX = 100# Return number of ways to which numbers# that are greater than given number can# be added to get sum.def numberofways(n, m): dp = [[0 for i in range(n+2)] for j in range(n+2)] dp[0][n + 1] = 1 # Filling the table. k is for numbers # greater than or equal that are allowed. for k in range(n, m-1, -1): # i is for sum for i in range(n + 1): # initializing dp[i][k] to number # ways to get sum using numbers # greater than or equal k+1 dp[i][k] = dp[i][k + 1] # if i > k if (i - k >= 0): dp[i][k] = (dp[i][k] + dp[i - k][k]) return dp[n][m]# Driver Codeif __name__ == "__main__": n, m = 3, 1 print(numberofways(n, m))# This code is contributed by Ryuga |
C#
// C# program to find number of ways to// which numbers that are greater than// given number can be added to get sum.using System;class GFG { // Return number of ways to which numbers // that are greater than given number can // be added to get sum. static int numberofways(int n, int m) { int[, ] dp = new int[n + 2, n + 2]; dp[0, n + 1] = 1; // Filling the table. k is for numbers // greater than or equal that are allowed. for (int k = n; k >= m; k--) { // i is for sum for (int i = 0; i <= n; i++) { // initializing dp[i][k] to number // ways to get sum using numbers // greater than or equal k+1 dp[i, k] = dp[i, k + 1]; // if i > k if (i - k >= 0) dp[i, k] = (dp[i, k] + dp[i - k, k]); } } return dp[n, m]; } // Driver Program public static void Main() { int n = 3, m = 1; Console.WriteLine(numberofways(n, m)); }}/*This code is contributed by vt_m.*/ |
PHP
<?php // PHP Program to find number of ways to// which numbers that are greater than// given number can be added to get sum.$MAX = 100;// Return number of ways to which numbers// that are greater than given number can// be added to get sum.function numberofways($n, $m){ global $MAX; $dp = array_fill(0, $n + 2, array_fill(0, $n+2, NULL)); $dp[0][$n + 1] = 1; // Filling the table. k is for numbers // greater than or equal that are allowed. for ($k = $n; $k >= $m; $k--) { // i is for sum for ($i = 0; $i <= $n; $i++) { // initializing dp[i][k] to number // ways to get sum using numbers // greater than or equal k+1 $dp[$i][$k] = $dp[$i][$k + 1]; // if i > k if ($i - $k >= 0) $dp[$i][$k] = ($dp[$i][$k] + $dp[$i - $k][$k]); } } return $dp[$n][$m];} // Driver Program $n = 3; $m = 1; echo numberofways($n, $m) ; return 0; // This code is contributed by ChitraNayal?> |
Javascript
<script>// Javascript Program to find number of ways to// which numbers that are greater than// given number can be added to get sum. // Return number of ways to which numbers // that are greater than given number can // be added to get sum. function numberofways(n,m) { let dp=new Array(n+2); for(let i=0;i<dp.length;i++) { dp[i]=new Array(n+2); for(let j=0;j<dp[i].length;j++) { dp[i][j]=0; } } dp[0][n + 1] = 1; // Filling the table. k is for numbers // greater than or equal that are allowed. for (let k = n; k >= m; k--) { // i is for sum for (let i = 0; i <= n; i++) { // initializing dp[i][k] to number // ways to get sum using numbers // greater than or equal k+1 dp[i][k] = dp[i][k + 1]; // if i > k if (i - k >= 0) dp[i][k] = (dp[i][k] + dp[i - k][k]); } } return dp[n][m]; } // Driver Program let n = 3, m = 1; document.write(numberofways(n, m)); // This code is contributed by avanitrachhadiya2155</script> |
Output
3
Time Complexity: O((n-m)*n)
Auxiliary Space: O(n*n)
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