# How to print maximum number of A’s using given four keys

This is a famous interview question asked in Google, Paytm and many other company interviews.

Below is the problem statement.

Imagine you have a special keyboard with the following keys: Key 1: Prints 'A' on screen Key 2: (Ctrl-A): Select screen Key 3: (Ctrl-C): Copy selection to buffer Key 4: (Ctrl-V): Print buffer on screen appending it after what has already been printed. If you can only press the keyboard for N times (with the above four keys), write a program to produce maximum numbers of A's. That is to say, the input parameter is N (No. of keys that you can press), the output is M (No. of As that you can produce).

**Examples:**

Input: N = 3 Output: 3 We can at most get 3 A's on screen by pressing following key sequence. A, A, A Input: N = 7 Output: 9 We can at most get 9 A's on screen by pressing following key sequence. A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V Input: N = 11 Output: 27 We can at most get 27 A's on screen by pressing following key sequence. A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V, Ctrl A, Ctrl C, Ctrl V, Ctrl V

Below are few important points to note.

a) For N < 7, the output is N itself.

b) Ctrl V can be used multiple times to print current buffer (See last two examples above). The idea is to compute the optimal string length for N keystrokes by using a simple insight. The sequence of N keystrokes which produces an optimal string length will end with a suffix of Ctrl-A, a Ctrl-C, followed by only Ctrl-V’s . (For N > 6)

The task is to find out the break=point after which we get the above suffix of keystrokes. Definition of a breakpoint is that instance after which we need to only press Ctrl-A, Ctrl-C once and the only Ctrl-V’s afterwards to generate the optimal length. If we loop from N-3 to 1 and choose each of these values for the break-point, and compute that optimal string they would produce. Once the loop ends, we will have the maximum of the optimal lengths for various breakpoints, thereby giving us the optimal length for N keystrokes.

Below is implementation based on above idea.

## C

`/* A recursive C program to print maximum number of A's using ` ` ` `following four keys */` `#include <stdio.h> ` ` ` `// A recursive function that returns the optimal length string ` `// for N keystrokes ` `int` `findoptimal(` `int` `N) ` `{ ` ` ` `// The optimal string length is N when N is smaller than 7 ` ` ` `if` `(N <= 6) ` ` ` `return` `N; ` ` ` ` ` `// Initialize result ` ` ` `int` `max = 0; ` ` ` ` ` `// TRY ALL POSSIBLE BREAK-POINTS ` ` ` `// For any keystroke N, we need to loop from N-3 keystrokes ` ` ` `// back to 1 keystroke to find a breakpoint 'b' after which we ` ` ` `// will have Ctrl-A, Ctrl-C and then only Ctrl-V all the way. ` ` ` `int` `b; ` ` ` `for` `(b = N - 3; b >= 1; b--) { ` ` ` `// If the breakpoint is s at b'th keystroke then ` ` ` `// the optimal string would have length ` ` ` `// (n-b-1)*screen[b-1]; ` ` ` `int` `curr = (N - b - 1) * findoptimal(b); ` ` ` `if` `(curr > max) ` ` ` `max = curr; ` ` ` `} ` ` ` `return` `max; ` `} ` ` ` `// Driver program ` `int` `main() ` `{ ` ` ` `int` `N; ` ` ` ` ` `// for the rest of the array we will rely on the previous ` ` ` `// entries to compute new ones ` ` ` `for` `(N = 1; N <= 20; N++) ` ` ` `printf` `(` `"Maximum Number of A's with %d keystrokes is %d\n"` `, ` ` ` `N, findoptimal(N)); ` `} ` |

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## Java

`/* A recursive Java program to print ` ` ` `maximum number of A's using ` ` ` `following four keys */` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` `// A recursive function that returns ` ` ` `// the optimal length string for N keystrokes ` ` ` `static` `int` `findoptimal(` `int` `N) ` ` ` `{ ` ` ` `// The optimal string length is N ` ` ` `// when N is smaller than 7 ` ` ` `if` `(N <= ` `6` `) ` ` ` `return` `N; ` ` ` ` ` `// Initialize result ` ` ` `int` `max = ` `0` `; ` ` ` ` ` `// TRY ALL POSSIBLE BREAK-POINTS ` ` ` `// For any keystroke N, we need to ` ` ` `// loop from N-3 keystrokes back to ` ` ` `// 1 keystroke to find a breakpoint ` ` ` `// 'b' after which we will have Ctrl-A, ` ` ` `// Ctrl-C and then only Ctrl-V all the way. ` ` ` `int` `b; ` ` ` `for` `(b = N - ` `3` `; b >= ` `1` `; b--) { ` ` ` `// If the breakpoint is s at b'th ` ` ` `// keystroke then the optimal string ` ` ` `// would have length ` ` ` `// (n-b-1)*screen[b-1]; ` ` ` `int` `curr = (N - b - ` `1` `) * findoptimal(b); ` ` ` `if` `(curr > max) ` ` ` `max = curr; ` ` ` `} ` ` ` `return` `max; ` ` ` `} ` ` ` ` ` `// Driver program ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `N; ` ` ` ` ` `// for the rest of the array we ` ` ` `// will rely on the previous ` ` ` `// entries to compute new ones ` ` ` `for` `(N = ` `1` `; N <= ` `20` `; N++) ` ` ` `System.out.println(` `"Maximum Number of A's with keystrokes is "` `+ N + findoptimal(N)); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

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## Python3

`# A recursive Python3 program to print maximum ` `# number of A's using following four keys ` ` ` `# A recursive function that returns ` `# the optimal length string for N keystrokes ` ` ` `def` `findoptimal(N): ` ` ` ` ` `# The optimal string length is ` ` ` `# N when N is smaller than ` ` ` `if` `N<` `=` `6` `: ` ` ` `return` `N ` ` ` ` ` `# Initialize result ` ` ` `maxi ` `=` `0` ` ` ` ` `# TRY ALL POSSIBLE BREAK-POINTS ` ` ` `# For any keystroke N, we need ` ` ` `# to loop from N-3 keystrokes ` ` ` `# back to 1 keystroke to find ` ` ` `# a breakpoint 'b' after which we ` ` ` `# will have Ctrl-A, Ctrl-C and then ` ` ` `# only Ctrl-V all the way. ` ` ` `for` `b ` `in` `range` `(N` `-` `3` `, ` `0` `, ` `-` `1` `): ` ` ` `curr ` `=` `(N` `-` `b` `-` `1` `)` `*` `findoptimal(b) ` ` ` `if` `curr>maxi: ` ` ` `maxi ` `=` `curr ` ` ` ` ` `return` `maxi ` `# Driver program ` `if` `__name__` `=` `=` `'__main__'` `: ` ` ` ` ` `# for the rest of the array we will ` `# rely on the previous ` `# entries to compute new ones ` ` ` `for` `n ` `in` `range` `(` `1` `, ` `21` `): ` ` ` `print` `(` `'Maximum Number of As with '` `, n, ` `'keystrokes is '` `, findoptimal(n)) ` ` ` `# this code is contibuted by sahilshelangia ` |

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## C#

`/* A recursive C# program ` `to print maximum number ` `of A's using following ` `four keys */` `using` `System; ` ` ` `class` `GFG { ` ` ` `// A recursive function that ` ` ` `// returns the optimal length ` ` ` `// string for N keystrokes ` ` ` `static` `int` `findoptimal(` `int` `N) ` ` ` `{ ` ` ` `// The optimal string length ` ` ` `// is N when N is smaller than 7 ` ` ` `if` `(N <= 6) ` ` ` `return` `N; ` ` ` ` ` `// Initialize result ` ` ` `int` `max = 0; ` ` ` ` ` `// TRY ALL POSSIBLE BREAK-POINTS ` ` ` `// For any keystroke N, we need ` ` ` `// to loop from N-3 keystrokes ` ` ` `// back to 1 keystroke to find ` ` ` `// a breakpoint 'b' after which ` ` ` `// we will have Ctrl-A, Ctrl-C ` ` ` `// and then only Ctrl-V all the way. ` ` ` `int` `b; ` ` ` `for` `(b = N - 3; b >= 1; b--) { ` ` ` `// If the breakpoint is s ` ` ` `// at b'th keystroke then ` ` ` `// the optimal string would ` ` ` `// have length (n-b-1)*screen[b-1]; ` ` ` `int` `curr = (N - b - 1) * findoptimal(b); ` ` ` `if` `(curr > max) ` ` ` `max = curr; ` ` ` `} ` ` ` `return` `max; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `N; ` ` ` ` ` `// for the rest of the array ` ` ` `// we will rely on the ` ` ` `// previous entries to compute ` ` ` `// new ones ` ` ` `for` `(N = 1; N <= 20; N++) ` ` ` `Console.WriteLine(` `"Maximum Number of A's with "` `+ N + ` `" keystrokes is "` `+ findoptimal(N)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007 ` |

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## PHP

`<?php ` `/* A recursive PHP program to print maximum number of A's using ` `following four keys */` ` ` `// A recursive function that returns the optimal ` `// length string for N keystrokes ` `function` `findoptimal(` `$N` `) ` `{ ` ` ` `// The optimal string length is ` ` ` `// N when N is smaller than 7 ` ` ` `if` `(` `$N` `<= 6) ` ` ` `return` `$N` `; ` ` ` ` ` `// Initialize result ` ` ` `$max` `= 0; ` ` ` ` ` `// TRY ALL POSSIBLE BREAK-POINTS ` ` ` `// For any keystroke N, we need to loop from N-3 keystrokes ` ` ` `// back to 1 keystroke to find a breakpoint 'b' after which we ` ` ` `// will have Ctrl-A, Ctrl-C and then only Ctrl-V all the way. ` ` ` `$b` `; ` ` ` `for` `(` `$b` `= ` `$N` `- 3; ` `$b` `>= 1; ` `$b` `-= 1) ` ` ` `{ ` ` ` `// If the breakpoint is s at b'th keystroke then ` ` ` `// the optimal string would have length ` ` ` `// (n-b-1)*screen[b-1]; ` ` ` `$curr` `= (` `$N` `- ` `$b` `- 1) * findoptimal(` `$b` `); ` ` ` `if` `(` `$curr` `> ` `$max` `) ` ` ` `$max` `= ` `$curr` `; ` ` ` `} ` ` ` `return` `$max` `; ` `} ` ` ` `// Driver code ` `$N` `; ` ` ` `// for the rest of the array we will rely on the previous ` `// entries to compute new ones ` `for` `(` `$N` `= 1; ` `$N` `<= 20; ` `$N` `+= 1) ` ` ` `echo` `(` `"Maximum Number of A's with"` ` ` `.` `$N` `.` `"keystrokes is "` `.findoptimal(` `$N` `).` `"\n"` `); ` ` ` `// This code is contributed by aman neekhara ` `?> ` |

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**Output:**

Maximum Number of A's with 1 keystrokes is 1 Maximum Number of A's with 2 keystrokes is 2 Maximum Number of A's with 3 keystrokes is 3 Maximum Number of A's with 4 keystrokes is 4 Maximum Number of A's with 5 keystrokes is 5 Maximum Number of A's with 6 keystrokes is 6 Maximum Number of A's with 7 keystrokes is 9 Maximum Number of A's with 8 keystrokes is 12 Maximum Number of A's with 9 keystrokes is 16 Maximum Number of A's with 10 keystrokes is 20 Maximum Number of A's with 11 keystrokes is 27 Maximum Number of A's with 12 keystrokes is 36 Maximum Number of A's with 13 keystrokes is 48 Maximum Number of A's with 14 keystrokes is 64 Maximum Number of A's with 15 keystrokes is 81 Maximum Number of A's with 16 keystrokes is 108 Maximum Number of A's with 17 keystrokes is 144 Maximum Number of A's with 18 keystrokes is 192 Maximum Number of A's with 19 keystrokes is 256 Maximum Number of A's with 20 keystrokes is 324

The above function computes the same subproblems again and again. Recomputations of same subproblems can be avoided by storing the solutions to subproblems and solving problems in bottom up manner.

Below is Dynamic Programming based C implementation where an auxiliary array screen[N] is used to store result of subproblems.

## C

`/* A Dynamic Programming based C program to find maximum number of A's ` ` ` `that can be printed using four keys */` `#include <stdio.h> ` ` ` `// this function returns the optimal length string for N keystrokes ` `int` `findoptimal(` `int` `N) ` `{ ` ` ` `// The optimal string length is N when N is smaller than 7 ` ` ` `if` `(N <= 6) ` ` ` `return` `N; ` ` ` ` ` `// An array to store result of subproblems ` ` ` `int` `screen[N]; ` ` ` ` ` `int` `b; ` `// To pick a breakpoint ` ` ` ` ` `// Initializing the optimal lengths array for uptil 6 input ` ` ` `// strokes. ` ` ` `int` `n; ` ` ` `for` `(n = 1; n <= 6; n++) ` ` ` `screen[n - 1] = n; ` ` ` ` ` `// Solve all subproblems in bottom manner ` ` ` `for` `(n = 7; n <= N; n++) { ` ` ` `// Initialize length of optimal string for n keystrokes ` ` ` `screen[n - 1] = 0; ` ` ` ` ` `// For any keystroke n, we need to loop from n-3 keystrokes ` ` ` `// back to 1 keystroke to find a breakpoint 'b' after which we ` ` ` `// will have ctrl-a, ctrl-c and then only ctrl-v all the way. ` ` ` `for` `(b = n - 3; b >= 1; b--) { ` ` ` `// if the breakpoint is at b'th keystroke then ` ` ` `// the optimal string would have length ` ` ` `// (n-b-1)*screen[b-1]; ` ` ` `int` `curr = (n - b - 1) * screen[b - 1]; ` ` ` `if` `(curr > screen[n - 1]) ` ` ` `screen[n - 1] = curr; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `screen[N - 1]; ` `} ` ` ` `// Driver program ` `int` `main() ` `{ ` ` ` `int` `N; ` ` ` ` ` `// for the rest of the array we will rely on the previous ` ` ` `// entries to compute new ones ` ` ` `for` `(N = 1; N <= 20; N++) ` ` ` `printf` `(` `"Maximum Number of A's with %d keystrokes is %d\n"` `, ` ` ` `N, findoptimal(N)); ` `} ` |

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## Java

`/* A Dynamic Programming based C ` ` ` `program to find maximum number ` ` ` `of A's that can be printed using ` ` ` `four keys */` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` `// this function returns the optimal ` ` ` `// length string for N keystrokes ` ` ` `static` `int` `findoptimal(` `int` `N) ` ` ` `{ ` ` ` `// The optimal string length is N ` ` ` `// when N is smaller than 7 ` ` ` `if` `(N <= ` `6` `) ` ` ` `return` `N; ` ` ` ` ` `// An array to store result ` ` ` `// of subproblems ` ` ` `int` `screen[] = ` `new` `int` `[N]; ` ` ` ` ` `int` `b; ` `// To pick a breakpoint ` ` ` ` ` `// Initializing the optimal lengths ` ` ` `// array for uptil 6 input strokes ` ` ` `int` `n; ` ` ` `for` `(n = ` `1` `; n <= ` `6` `; n++) ` ` ` `screen[n - ` `1` `] = n; ` ` ` ` ` `// Solve all subproblems in bottom manner ` ` ` `for` `(n = ` `7` `; n <= N; n++) { ` ` ` `// Initialize length of optimal ` ` ` `// string for n keystrokes ` ` ` `screen[n - ` `1` `] = ` `0` `; ` ` ` ` ` `// For any keystroke n, we need ` ` ` `// to loop from n-3 keystrokes ` ` ` `// back to 1 keystroke to find ` ` ` `// a breakpoint 'b' after which we ` ` ` `// will have ctrl-a, ctrl-c and ` ` ` `// then only ctrl-v all the way. ` ` ` `for` `(b = n - ` `3` `; b >= ` `1` `; b--) { ` ` ` `// if the breakpoint is ` ` ` `// at b'th keystroke then ` ` ` `// the optimal string would ` ` ` `// have length ` ` ` `// (n-b-1)*screen[b-1]; ` ` ` `int` `curr = (n - b - ` `1` `) * screen[b - ` `1` `]; ` ` ` `if` `(curr > screen[n - ` `1` `]) ` ` ` `screen[n - ` `1` `] = curr; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `screen[N - ` `1` `]; ` ` ` `} ` ` ` ` ` `// Driver program ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `N; ` ` ` ` ` `// for the rest of the array we will rely on the previous ` ` ` `// entries to compute new ones ` ` ` `for` `(N = ` `1` `; N <= ` `20` `; N++) ` ` ` `System.out.println(` `"Maximum Number of A's with keystrokes is "` `+ N + findoptimal(N)); ` ` ` `} ` `} ` ` ` `// This article is contributed by vt_m. ` |

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## Python3

`# A Dynamic Programming based Python program ` `# to find maximum number of A's ` `# that can be printed using four keys ` ` ` `# this function returns the optimal ` `# length string for N keystrokes ` `def` `findoptimal(N): ` ` ` ` ` `# The optimal string length is ` ` ` `# N when N is smaller than 7 ` ` ` `if` `(N <` `=` `6` `): ` ` ` `return` `N ` ` ` ` ` `# An array to store result of ` ` ` `# subproblems ` ` ` `screen ` `=` `[` `0` `]` `*` `N ` ` ` ` ` `# Initializing the optimal lengths ` ` ` `# array for uptil 6 input ` ` ` `# strokes. ` ` ` ` ` `for` `n ` `in` `range` `(` `1` `, ` `7` `): ` ` ` `screen[n` `-` `1` `] ` `=` `n ` ` ` ` ` `# Solve all subproblems in bottom manner ` ` ` `for` `n ` `in` `range` `(` `7` `, N ` `+` `1` `): ` ` ` ` ` `# Initialize length of optimal ` ` ` `# string for n keystrokes ` ` ` `screen[n` `-` `1` `] ` `=` `0` ` ` ` ` `# For any keystroke n, we need to ` ` ` `# loop from n-3 keystrokes ` ` ` `# back to 1 keystroke to find a breakpoint ` ` ` `# 'b' after which we ` ` ` `# will have ctrl-a, ctrl-c and then only ` ` ` `# ctrl-v all the way. ` ` ` `for` `b ` `in` `range` `(n` `-` `3` `, ` `0` `, ` `-` `1` `): ` ` ` ` ` `# if the breakpoint is at b'th keystroke then ` ` ` `# the optimal string would have length ` ` ` `# (n-b-1)*screen[b-1]; ` ` ` `curr ` `=` `(n` `-` `b` `-` `1` `)` `*` `screen[b` `-` `1` `] ` ` ` `if` `(curr > screen[n` `-` `1` `]): ` ` ` `screen[n` `-` `1` `] ` `=` `curr ` ` ` ` ` `return` `screen[N` `-` `1` `] ` ` ` `# Driver program ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `# for the rest of the array we ` ` ` `# will rely on the previous ` ` ` `# entries to compute new ones ` ` ` `for` `N ` `in` `range` `(` `1` `, ` `21` `): ` ` ` `print` `(` `"Maximum Number of A's with "` `, N, ` `" keystrokes is "` `, ` ` ` `findoptimal(N)) ` ` ` `# this code is contributed by ` `# ChitraNayal ` |

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## C#

`/* A Dynamic Programming based C# ` `program to find maximum number ` `of A's that can be printed using ` `four keys */` `using` `System; ` `public` `class` `GFG { ` ` ` ` ` `// this function returns the optimal ` ` ` `// length string for N keystrokes ` ` ` `static` `int` `findoptimal(` `int` `N) ` ` ` `{ ` ` ` `// The optimal string length is N ` ` ` `// when N is smaller than 7 ` ` ` `if` `(N <= 6) ` ` ` `return` `N; ` ` ` ` ` `// An array to store result ` ` ` `// of subproblems ` ` ` `int` `[] screen = ` `new` `int` `[N]; ` ` ` ` ` `int` `b; ` `// To pick a breakpoint ` ` ` ` ` `// Initializing the optimal lengths ` ` ` `// array for uptil 6 input strokes ` ` ` `int` `n; ` ` ` `for` `(n = 1; n <= 6; n++) ` ` ` `screen[n - 1] = n; ` ` ` ` ` `// Solve all subproblems in bottom manner ` ` ` `for` `(n = 7; n <= N; n++) { ` ` ` `// Initialize length of optimal ` ` ` `// string for n keystrokes ` ` ` `screen[n - 1] = 0; ` ` ` ` ` `// For any keystroke n, we need ` ` ` `// to loop from n-3 keystrokes ` ` ` `// back to 1 keystroke to find ` ` ` `// a breakpoint 'b' after which we ` ` ` `// will have ctrl-a, ctrl-c and ` ` ` `// then only ctrl-v all the way. ` ` ` `for` `(b = n - 3; b >= 1; b--) { ` ` ` `// if the breakpoint is ` ` ` `// at b'th keystroke then ` ` ` `// the optimal string would ` ` ` `// have length ` ` ` `// (n-b-1)*screen[b-1]; ` ` ` `int` `curr = (n - b - 1) * screen[b - 1]; ` ` ` `if` `(curr > screen[n - 1]) ` ` ` `screen[n - 1] = curr; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `screen[N - 1]; ` ` ` `} ` ` ` ` ` `// Driver program ` ` ` `public` `static` `void` `Main(String[] args) ` ` ` `{ ` ` ` `int` `N; ` ` ` ` ` `// for the rest of the array we will rely on the previous ` ` ` `// entries to compute new ones ` ` ` `for` `(N = 1; N <= 20; N++) ` ` ` `Console.WriteLine(` `"Maximum Number of A's with {0} keystrokes is {1}\n"` `, ` ` ` `N, findoptimal(N)); ` ` ` `} ` `} ` ` ` `// This code is contributed by PrinciRaj1992 ` |

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Output:

Maximum Number of A's with 1 keystrokes is 1 Maximum Number of A's with 2 keystrokes is 2 Maximum Number of A's with 3 keystrokes is 3 Maximum Number of A's with 4 keystrokes is 4 Maximum Number of A's with 5 keystrokes is 5 Maximum Number of A's with 6 keystrokes is 6 Maximum Number of A's with 7 keystrokes is 9 Maximum Number of A's with 8 keystrokes is 12 Maximum Number of A's with 9 keystrokes is 16 Maximum Number of A's with 10 keystrokes is 20 Maximum Number of A's with 11 keystrokes is 27 Maximum Number of A's with 12 keystrokes is 36 Maximum Number of A's with 13 keystrokes is 48 Maximum Number of A's with 14 keystrokes is 64 Maximum Number of A's with 15 keystrokes is 81 Maximum Number of A's with 16 keystrokes is 108 Maximum Number of A's with 17 keystrokes is 144 Maximum Number of A's with 18 keystrokes is 192 Maximum Number of A's with 19 keystrokes is 256 Maximum Number of A's with 20 keystrokes is 324

Thanks to **Gaurav Saxena** for providing the above approach to solve this problem.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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