codility MinMaxDivision

Task description

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:

int solution(int K, int M, vector<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.

#include <algorithm>
#include <numeric>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int check(vector<int> &A, int max_sum);

int solution(int K, int M, vector<int> &A) {
    // write your code in C++11
    int right = accumulate(A.begin(), A.end(), 0);
    int left  = *max_element(A.begin(), A.end());
    
    if(K == 1)          return right;
    if(K >= A.size())   return left;
    
    int ret = 0; 
    while(left <= right){
        int middle = ((right-left)>>1) + left;
        if(check(A, middle) <= K){
            right = middle-1;
            ret = middle;
        }
        else{
            left = middle+1;
        }
    }
    
    return ret;
}

int check(vector<int> &A, int max_sum)
{
    int size = A.size();
    int min_groups = 1;
    int current_sum = 0;
    
    for(int i = 0; i < size; i++){
        if(current_sum <= max_sum - A[i]){
            current_sum += A[i];
        }
        else{
            current_sum = A[i];
            min_groups++;
        }
    }
    
    return min_groups;
}