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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Manasa and Stones
  5. Discussions

Manasa and Stones

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950 Discussions

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  • chauhansatish978
     + 0 comments
    int diff = abs(b-a);
    int ans = (n-1)*min(a,b);
    
    vector<int>result;
    result.push_back(ans);
    
    for(int i = 0; i< n-1;i++){
        if(result[i]+diff == result[i]) continue;
        result.push_back(result[i]+diff);  
    } 
    
    return  result;

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/BXNoCVP0XEE Solution 1 : 

    vector<int> stones(int n, int a, int b) {
    if(a > b){
        a = b + a;
        b = a - b;
        a = a - b;
    }
    int step = b - a;
    vector<int> result;
    for(int i = a * (n-1); i < b * (n-1); i += step) result.push_back(i);
    result.push_back(b * (n-1));
    return result;
}

    Solution 2 :

    vector<int> stones(int n, int a, int b) {
    vector<int> result;
    for(int i = min(a, b) * (n-1); i < max(a,b) * (n-1); i += abs(b - a)) result.push_back(i);
    result.push_back(max(a, b) * (n-1));
    return result;
}

  • pub_ashishk28
     + 0 comments
    def stones(n, a, b):
    # Write your code here
    x = [ sum(i) for i in combinations_with_replacement([a,b], n -1 )]
        
    return sorted(set(x))

  • dorirgob
     + 0 comments
     public static List<Integer> stones(int n, int a, int b) {
    // Write your code here
    int bigger = (a >= b) ? a : b;
    int smaller = (a <= b) ? a : b;
    int diff = bigger - smaller;
    int min = smaller * (n - 1);
    int current = min;
    List<Integer> lastValues = new ArrayList<Integer>();
    
    if(diff == 0) {
      lastValues.add(current);
      return lastValues;
    }
    
    for(int i = 0; i < n; i++) {
      //System.out.println(current);
      lastValues.add(current);
      current += diff;
      

    }

return lastValues;

}

  • tangjiaqing
     + 0 comments

    I think there is something wrong with the examples except for the first one. The first parameter represents non-zero stone. However, in the rest example, including the test cases, n represents total number of stones including non-zero. This is the solution to pass all of the tests.

    public static List<int> Run(int count, int diff1, int diff2)
{
	if (diff1 == diff2) return [diff1 * (count - 1)];

	var small = diff1 < diff2 ? diff1 : diff2;
	var large = diff1 < diff2 ? diff2 : diff1;

	var start = small * (count - 1);
	var end = large * (count - 1);
	var step = large - small;

	if (start == 0) start += step;

	var faces = new List<int>();

	for (var i = start; i <= end; i += step)
	{
		faces.Add(i);
	}

	return faces;
}