Ways to arrange Balls such that adjacent balls are of different types | C ++ program to count number of ways to arrange three types of balls such that no two balls of same color are adjacent to each other ; table to store to store results of subproblems ; Returns count of arrangements where last placed ball is ' last ' . ' last ' is 0 for ' p ' , 1 for ' q ' and 2 for ' r ' ; if number of balls of any color becomes less than 0 the number of ways arrangements is 0. ; If last ball required is of type P and the number of balls of P type is 1 while number of balls of other color is 0 the number of ways is 1. ; Same case as above for ' q ' and ' r ' ; If this subproblem is already evaluated ; if last ball required is P and the number of ways is the sum of number of ways to form sequence with ' p - 1' P balls , q Q Balls and r R balls ending with Q and R . ; Same as above case for ' q ' and ' r ' ; else ( last == 2 ) ; Returns count of required arrangements ; Initialize ' dp ' array ; Three cases arise : return countWays ( p , q , r , 0 ) + Last required balls is type P countWays ( p , q , r , 1 ) + Last required balls is type Q countWays ( p , q , r , 2 ) ; Last required balls is type R ; Driver code to test above
#include <bits/stdc++.h> NEW_LINE using namespace std ; #define MAX  100 NEW_LINE int dp [ MAX ] [ MAX ] [ MAX ] [ 3 ] ; int countWays ( int p , int q , int r , int last ) { if ( p < 0 q < 0 r < 0 ) return 0 ; if ( p == 1 && q == 0 && r == 0 && last == 0 ) return 1 ; if ( p == 0 && q == 1 && r == 0 && last == 1 ) return 1 ; if ( p == 0 && q == 0 && r == 1 && last == 2 ) return 1 ; if ( dp [ p ] [ q ] [ r ] [ last ] != -1 ) return dp [ p ] [ q ] [ r ] [ last ] ; if ( last == 0 ) dp [ p ] [ q ] [ r ] [ last ] = countWays ( p - 1 , q , r , 1 ) + countWays ( p - 1 , q , r , 2 ) ; else if ( last == 1 ) dp [ p ] [ q ] [ r ] [ last ] = countWays ( p , q - 1 , r , 0 ) + countWays ( p , q - 1 , r , 2 ) ; dp [ p ] [ q ] [ r ] [ last ] = countWays ( p , q , r - 1 , 0 ) + countWays ( p , q , r - 1 , 1 ) ; return dp [ p ] [ q ] [ r ] [ last ] ; } int countUtil ( int p , int q , int r ) { memset ( dp , -1 , sizeof ( dp ) ) ; } int main ( ) { int p = 1 , q = 1 , r = 1 ; printf ( " % d " , countUtil ( p , q , r ) ) ; return 0 ; }