Codility - Lesson 14 Binary Search Algorithm - 1. MinMaxDivision

Source Link:
https://codility.com/programmers/lessons/14-binary_search_algorithm/min_max_division/

Question:

You are given integers K, M and a non-empty zero-indexed array A consisting of N integers. Every element of the array is not greater than M. You should divide this array into K blocks of consecutive elements. The size of the block is any integer between 0 and N. Every element of the array should belong to some block. The sum of the block from X to Y equals A[X] + A[X + 1] + ... + A[Y]. The sum of empty block equals 0. The large sum is the maximal sum of any block.

For example, you are given integers K = 3, M = 5 and array A such that:
A[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2 A[6] = 2  

The array can be divided, for example, into the following blocks:

• [2, 1, 5, 1, 2, 2, 2], [], [] with a large sum of 15;
• [2], [1, 5, 1, 2], [2, 2] with a large sum of 9;
• [2, 1, 5], [], [1, 2, 2, 2] with a large sum of 8;
• [2, 1], [5, 1], [2, 2, 2] with a large sum of 6.

The goal is to minimize the large sum. In the above example, 6 is the minimal large sum.

Write a function:
class Solution { public int solution(int K, int M, int[] A); }
that, given integers K, M and a non-empty zero-indexed array A consisting of N integers, returns the minimal large sum.

For example, given K = 3, M = 5 and array A such that:
A[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2 A[6] = 2  
the function should return 6, as explained above.

Assume that:

• N and K are integers within the range [1..100,000];
• M is an integer within the range [0..10,000];
• each element of array A is an integer within the range [0..M].

Complexity:

• expected worst-case time complexity is O(N*log(N+M));
• expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

My Solution:
Notes:
1. main idea:
// The goal is to find the "minimal large sum"
// We use "binary search" to find it (so, it can be fast)
// We assume that the "min max Sum" will be 
// between "min" and "max", each time we try "mid"   
2. 
for(int i=0; i<A.length; i++){
maxSum = maxSum + A[i]; // maxSum: sum of all elements
minSum = Math.max(minSum, A[i]); // minSum: at least one max element 
}
3. the max one must be an "ok" result
int possibleResult = maxSum; 
4. do "binary search" (search for better Sum)
while(minSum <= maxSum){
int midSum = (minSum + maxSum) /2; 
boolean ok = checkDivisable(midSum, K, A);
if(ok == true){
possibleResult = midSum;
maxSum = midSum - 1; 
}
else{ 
minSum = midSum + 1;
}
}
5.
return possibleResult;
6. check if it can be divided by using the minMaxSum = "mid", into K blocks ?
public boolean checkDivisable(int mid, int k, int[] a)