Maximum number of segments of lengths a, b and c
Given a positive integer N, find the maximum number of segments of lengths a, b and c that can be formed from N .
Examples :
Input : N = 7, a = 5, b, = 2, c = 5 Output : 2 N can be divided into 2 segments of lengths 2 and 5. For the second example, Input : N = 17, a = 2, b = 1, c = 3 Output : 17 N can be divided into 17 segments of 1 or 8 segments of 2 and 1 segment of 1. But 17 segments of 1 is greater than 9 segments of 2 and 1.
Approach : The approach used is Dynamic Programming. The base dp0 will be 0 as initially it has no segments. After that, iterate from 1 to n, and for each of the 3 states i.e, dpi+a, dpi+b and dpi+c, store the maximum value obtained by either using or not using the a, b or c segment.
The 3 states to deal with are :
dpi+a=max(dpi+1, dpi+a);
dpi+b=max(dpi+1, dpi+b);
dpi+c=max(dpi+1, dpi+c);
Below is the implementation of above idea :
C++
// C++ implementation to divide N into// maximum number of segments// of length a, b and c#include <bits/stdc++.h>using namespace std;// function to find the maximum// number of segmentsint maximumSegments(int n, int a, int b, int c){ // stores the maximum number of // segments each index can have int dp[n + 1]; // initialize with -1 memset(dp, -1, sizeof(dp)); // 0th index will have 0 segments // base case dp[0] = 0; // traverse for all possible // segments till n for (int i = 0; i < n; i++) { if (dp[i] != -1) { // conditions if(i + a <= n ) //avoid buffer overflow dp[i + a] = max(dp[i] + 1, dp[i + a]); if(i + b <= n ) //avoid buffer overflow dp[i + b] = max(dp[i] + 1, dp[i + b]); if(i + c <= n ) //avoid buffer overflow dp[i + c] = max(dp[i] + 1, dp[i + c]); } } return dp[n];}// Driver codeint main(){ int n = 7, a = 5, b = 2, c = 5; cout << maximumSegments(n, a, b, c); return 0;} |
Java
// Java implementation to divide N into// maximum number of segments// of length a, b and cimport java.util.*;class GFG{ // function to find the maximum // number of segments static int maximumSegments(int n, int a, int b, int c) { // stores the maximum number of // segments each index can have int dp[] = new int[n + 10]; // initialize with -1 Arrays.fill(dp, -1); // 0th index will have 0 segments // base case dp[0] = 0; // traverse for all possible // segments till n for (int i = 0; i < n; i++) { if (dp[i] != -1) { // conditions if(i + a <= n ) //avoid buffer overflow dp[i + a] = Math.max(dp[i] + 1, dp[i + a]); if(i + b <= n ) //avoid buffer overflow dp[i + b] = Math.max(dp[i] + 1, dp[i + b]); if(i + c <= n ) //avoid buffer overflow dp[i + c] = Math.max(dp[i] + 1, dp[i + c]); } } return dp[n]; } // Driver code public static void main(String arg[]) { int n = 7, a = 5, b = 2, c = 5; System.out.print(maximumSegments(n, a, b, c)); }}// This code is contributed by Anant Agarwal. |
Python3
# Python implementation# to divide N into maximum# number of segments of# length a, b and c# function to find# the maximum number# of segmentsdef maximumSegments(n, a, b, c) : # stores the maximum # number of segments # each index can have dp = [-1] * (n + 10) # 0th index will have # 0 segments base case dp[0] = 0 # traverse for all possible # segments till n for i in range(0, n) : if (dp[i] != -1) : # conditions if(i + a <= n ): # avoid buffer overflow dp[i + a] = max(dp[i] + 1, dp[i + a]) if(i + b <= n ): # avoid buffer overflow dp[i + b] = max(dp[i] + 1, dp[i + b]) if(i + c <= n ): # avoid buffer overflow dp[i + c] = max(dp[i] + 1, dp[i + c]) return dp[n]# Driver coden = 7a = 5b = 2c = 5print (maximumSegments(n, a, b, c))# This code is contributed by# Manish Shaw(manishshaw1) |
C#
// C# implementation to divide N into// maximum number of segments// of length a, b and cusing System;class GFG{ // function to find the maximum // number of segments static int maximumSegments(int n, int a, int b, int c) { // stores the maximum number of // segments each index can have int []dp = new int[n + 10]; // initialize with -1 for(int i = 0; i < n + 10; i++) dp[i]= -1; // 0th index will have 0 segments // base case dp[0] = 0; // traverse for all possible // segments till n for (int i = 0; i < n; i++) { if (dp[i] != -1) { // conditions if(i + a <= n ) // avoid buffer overflow dp[i + a] = Math.Max(dp[i] + 1, dp[i + a]); if(i + b <= n ) // avoid buffer overflow dp[i + b] = Math.Max(dp[i] + 1, dp[i + b]); if(i + c <= n ) // avoid buffer overflow dp[i + c] = Math.Max(dp[i] + 1, dp[i + c]); } } return dp[n]; } // Driver code public static void Main() { int n = 7, a = 5, b = 2, c = 5; Console.Write(maximumSegments(n, a, b, c)); }}// This code is contributed by nitin mittal |
PHP
<?php// PHP implementation to divide// N into maximum number of// segments of length a, b and c// function to find the maximum// number of segmentsfunction maximumSegments($n, $a, $b, $c){ // stores the maximum // number of segments // each index can have $dp = array(); // initialize with -1 for($i = 0; $i < $n + 10; $i++) $dp[$i]= -1; // 0th index will have // 0 segments base case $dp[0] = 0; // traverse for all possible // segments till n for ($i = 0; $i < $n; $i++) { if ($dp[$i] != -1) { // conditions if($i + $a <= $n ) // avoid buffer overflow $dp[$i + $a] = max($dp[$i] + 1, $dp[$i + $a]); if($i + $b <= $n ) // avoid buffer overflow $dp[$i + $b] = max($dp[$i] + 1, $dp[$i + $b]); if($i + $c <= $n ) // avoid buffer overflow $dp[$i + $c] = max($dp[$i] + 1, $dp[$i + $c]); } } return $dp[$n];}// Driver code$n = 7; $a = 5;$b = 2; $c = 5;echo (maximumSegments($n, $a, $b, $c));// This code is contributed by// Manish Shaw(manishshaw1)?> |
Javascript
<script>// JavaScript program implementation to divide N into// maximum number of segments// of length a, b and c // function to find the maximum // number of segments function maximumSegments(n, a, b, c) { // stores the maximum number of // segments each index can have let dp = []; // initialize with -1 for(let i = 0; i < n + 10; i++) dp[i]= -1; // 0th index will have 0 segments // base case dp[0] = 0; // traverse for all possible // segments till n for (let i = 0; i < n; i++) { if (dp[i] != -1) { // conditions if(i + a <= n ) //avoid buffer overflow dp[i + a] = Math.max(dp[i] + 1, dp[i + a]); if(i + b <= n ) //avoid buffer overflow dp[i + b] = Math.max(dp[i] + 1, dp[i + b]); if(i + c <= n ) //avoid buffer overflow dp[i + c] = Math.max(dp[i] + 1, dp[i + c]); } } return dp[n]; } // Driver Code let n = 7, a = 5, b = 2, c = 5; document.write(maximumSegments(n, a, b, c));// This code is contributed by susmitakundugoaldanga.</script> |
Output :
2
Time complexity : O(n)
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