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 segments int 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 code int 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 c import 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 segments def 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 code n = 7 a = 5 b = 2 c = 5 print (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 c using 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 segments function 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) ?> |
Output :
2
Time complexity : O(n)
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