m-th summation of first n natural numbers is defined as following.
If m > 1 SUM(n, m) = SUM(SUM(n, m - 1), 1) Else SUM(n, 1) = Sum of first n natural numbers.
We are given m and n, we need to find SUM(n, m).
Examples:
Input : n = 4, m = 1
Output : SUM(4, 1) = 10
Explanation : 1 + 2 + 3 + 4 = 10
Input : n = 3, m = 2
Output : SUM(3, 2) = 21
Explanation : SUM(3, 2)
= SUM(SUM(3, 1), 1)
= SUM(6, 1)
= 21
Naive Approach : We can solve this problem using two nested loop, where outer loop iterate for m and inner loop iterate for n. After completion of a single outer iteration, we should update n as whole of inner loop got executed and value of n must be changed then. Time complexity should be O(n*m).
for (int i = 1;i <= m;i++)
{
sum = 0;
for (int j = 1;j <= n;j++)
sum += j;
n = sum; // update n
}
Efficient Approach :
We can use direct formula for sum of first n numbers to reduce time.
We can also use recursion. In this approach m = 1 will be our base condition and for any intermediate step SUM(n, m), we will call SUM (SUM(n, m-1), 1) and for a single step SUM(n, 1) = n * (n + 1) / 2 will be used. This will reduce our time complexity to O(m).
int SUM (int n, int m)
{
if (m == 1)
return (n * (n + 1) / 2);
int sum = SUM(n, m-1);
return (sum * (sum + 1) / 2);
}
Below is the implementation of above idea :
C++
// CPP program to find m-th summation #include <bits/stdc++.h>using namespace std;// Function to return mth summationint SUM(int n, int m){ // base case if (m == 1) return (n * (n + 1) / 2); int sum = SUM(n, m-1); return (sum * (sum + 1) / 2);}// driver programint main(){ int n = 5; int m = 3; cout << "SUM(" << n << ", " << m << "): " << SUM(n, m); return 0;} |
Java
// Java program to find m-th summation.class GFG { // Function to return mth summation static int SUM(int n, int m) { // base case if (m == 1) return (n * (n + 1) / 2); int sum = SUM(n, m - 1); return (sum * (sum + 1) / 2); } // Driver code public static void main(String[] args) { int n = 5; int m = 3; System.out.println("SUM(" + n + ", " + m + "): " + SUM(n, m)); }}// This code is contributed by Anant Agarwal. |
Python
# Python program to find m-th summation.def SUM(n, m): # base case if m == 1: return (n * (n + 1) // 2) su = SUM(n, m - 1) return (su * (su + 1) // 2)# Driver coden = 5m = 3print("SUM("+str(n)+", "+str(m)+"): "+str(SUM(n, m))) |
C#
// C# program to find m-th summation.using System;class GFG { // Function to return mth summation static int SUM(int n, int m) { // base case if (m == 1) return (n * (n + 1) / 2); int sum = SUM(n, m - 1); return (sum * (sum + 1) / 2); } // Driver Code public static void Main() { int n = 5; int m = 3; Console.Write("SUM(" + n + ", " + m + "): " + SUM(n, m)); }}// This code is contributed by Nitin Mittal. |
PHP
<?php// PHP program to find m-th summation // Function to return// mth summationfunction SUM($n, $m){ // base case if ($m == 1) return ($n * ($n + 1) / 2); $sum = SUM($n, $m - 1); return ($sum * ($sum + 1) / 2);} // Driver Code $n = 5; $m = 3; echo "SUM(" , $n , ", " , $m , "): " , SUM($n, $m);// This code is contributed by vt_m.?> |
Javascript
<script>// javascript program to find m-th summation // Function to return mth summationfunction SUM( n, m){ // base case if (m == 1) return (n * (n + 1) / 2); let sum = SUM(n, m-1); return (sum * (sum + 1) / 2);}// driver program let n = 5; let m = 3; document.write( "SUM(" + n + ", " + m + "): " + SUM(n, m));// This code contributed by Rajput-Ji </script> |
Output:
SUM(5, 3): 7260
Space complexity :- O(M)
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