Distirbute Candy - Finding minimum reproducible example of the problem

Question

The question is to distribute candies to N children.Each child has a rating. Distribution should be such that each child have at least one candy and children with higher rating gets more candies than their neighbors. Find minimum number of candies for such distribution.

Approach:

  iterate each child.
       if (rating of curr child < rating of next child)
              add to total and increase candy
       else if (rating of curr child equal to rating of next child)
              we can bring down candy for next child to 1
       else if ( rating of curr child > rating of next child)
              {     // if child ratings are {4 3 2 5} then 
                    //optimal is {3 2 1 2}
                  //the detailed code is below
              }

The detailed code is :

int n = A.size();
int count  = 0;
int step = 1;
int i = 0;
bool run = true;
while(run){
    if(i+1 ==n){
        count+=step;
        run = false;
    }
    else if(A[i+1] > A[i]){
        count+=step;
        step++;
        i+=1;;
    }else if(A[i+1] == A[i]){
        count+=step;
        step = 1;
      
        i+=1;
    }else {
        int j = i;
        while(A[i+1] < A[i]&& (i+1)<n){
            i+=1;
        }
        
        int x = i-j;
        step = max(step,x+1);
        count+=step;
        count+=((x*(x+1))/2);
        step = 2;
        if(i==n-1)
            run = false;
        i+=1;
        
    }
}


 return count;

The code doesn't produce expected output. Due to huge size of test case, I cannot determine cause of error. Can someone provide sample test case where it breaks?

The failed test case is attached in below link. The first number denotes size of array. Breaking test case

Just try small test cases until it breaks. Works most of the time in situations like this. — eesiraed, Jun 16 '20 at 17:20
Please start with the [tour] and read [ask]. You will have to provide test data and results inline but first reduce it to a [mcve]. As it stands, your question is off-topic. — Ulrich Eckhardt, Jun 16 '20 at 17:21

score 1 · Accepted Answer · answered Jun 17 '20 at 07:21

If you only need an example that exposes an error in your code, use 3 2 2 and stop reading.

I would suggest a very literal solution to the problem:

Initialize a result array of equal size as the A array with value 1 for each position (each child gets at least one candy)
Forward iterate the arrays and apply following logic
Backward iterate the arrays and apply following logic
Compute the sum of the result array

Logic for steps 2 and 3:

if (A[current] > A[previous] && result[current] <= result[previous])
    result[current] = result[previous] + 1;

score 0 · Answer 2 · answered Apr 17 '22 at 20:52

iterate each child. if (rating of curr child < rating of next child) add to total and increase candy else if (rating of curr child equal to rating of next child) we can bring down candy for next child to 1 else if ( rating of curr child > rating of next child) { // if child ratings are {4 3 2 5} then //optimal is {3 2 1 2} //the detailed

Distirbute Candy - Finding minimum reproducible example of the problem

Approach:

2 Answers2