Recursively break a number in 3 parts to get maximum sum - GeeksforGeeks
Given a number n, we can divide it in only three parts n/2, n/3 and n/4 (we will consider only integer part). The task is to find the maximum sum we can make by dividing number in three parts recursively and summing up them together.
A simple solution for this problem is to do it recursively. In each call we have to check only max((max(n/2) + max(n/3) + max(n/4)), n) and return it. Because either we can get maximum sum by breaking number in parts or number is itself maximum
An efficient solution for this problem is to use Dynamic programming because while breaking the number in parts recursively we have to perform some overlapping problems. For example part of n = 30 will be {15,10,7} and part of 15 will be {7,5,3} so we have to repeat the process for 7 two times for further breaking. To avoid this overlapping problem we are using Dynamic programming. We store values in an array and if for any number in recursive calls we have already solution for that number currently so we diretly extract it from array
Given a number n, we can divide it in only three parts n/2, n/3 and n/4 (we will consider only integer part). The task is to find the maximum sum we can make by dividing number in three parts recursively and summing up them together.
Input : n = 12
Output : 13
// We break n = 12 in three parts {12/2, 12/3, 12/4}
// = {6, 4, 3}, now current sum is = (6 + 4 + 3) = 13
// again we break 6 = {6/2, 6/3, 6/4} = {3, 2, 1} = 3 +
// 2 + 1 = 6 and further breaking 3, 2 and 1 we get maximum
// summation as 1, so breaking 6 in three parts produces
// maximum sum 6 only similarly breaking 4 in three
// parts we can get maximum sum 4 and same for 3 also.
// Thus maximum sum by breaking number in parts is=13
Input : n = 24
Output : 27
// We break n = 24 in three parts {24/2, 24/3, 24/4}
// = {12, 8, 6}, now current sum is = (12 + 8 + 6) = 26
// As seen in example, recursively breaking 12 would
// produce value 13. So our maximum sum is 13 + 8 + 6 = 27.
// Note that recursively breaking 8 and 6 doesn't produce
// more values, that is why they are not broken further.
A simple solution for this problem is to do it recursively. In each call we have to check only max((max(n/2) + max(n/3) + max(n/4)), n) and return it. Because either we can get maximum sum by breaking number in parts or number is itself maximum
