Query Kth Smallest Trimmed Number LeetCode Solution

Problem – Query Kth Smallest Trimmed Number LeetCode Solution

You are given a 0-indexed array of strings nums, where each string is of equal length and consists of only digits.

You are also given a 0-indexed 2D integer array queries where queries[i] = [ki, trimi]. For each queries[i], you need to:

  • Trim each number in nums to its rightmost trimi digits.
  • Determine the index of the kith smallest trimmed number in nums. If two trimmed numbers are equal, the number with the lower index is considered to be smaller.
  • Reset each number in nums to its original length.

Return an array answer of the same length as queries, where answer[i] is the answer to the ith query.

Note:

  • To trim to the rightmost x digits means to keep removing the leftmost digit, until only x digits remain.
  • Strings in nums may contain leading zeros.

Example 1:

Input: nums = ["102","473","251","814"], queries = [[1,1],[2,3],[4,2],[1,2]]
Output: [2,2,1,0]
Explanation:
1. After trimming to the last digit, nums = ["2","3","1","4"]. The smallest number is 1 at index 2.
2. Trimmed to the last 3 digits, nums is unchanged. The 2nd smallest number is 251 at index 2.
3. Trimmed to the last 2 digits, nums = ["02","73","51","14"]. The 4th smallest number is 73.
4. Trimmed to the last 2 digits, the smallest number is 2 at index 0.
   Note that the trimmed number "02" is evaluated as 2.

Example 2:

Input: nums = ["24","37","96","04"], queries = [[2,1],[2,2]]
Output: [3,0]
Explanation:
1. Trimmed to the last digit, nums = ["4","7","6","4"]. The 2nd smallest number is 4 at index 3.
   There are two occurrences of 4, but the one at index 0 is considered smaller than the one at index 3.
2. Trimmed to the last 2 digits, nums is unchanged. The 2nd smallest number is 24.

Constraints:

  • 1 <= nums.length <= 100
  • 1 <= nums[i].length <= 100
  • nums[i] consists of only digits.
  • All nums[i].length are equal.
  • 1 <= queries.length <= 100
  • queries[i].length == 2
  • 1 <= ki <= nums.length
  • 1 <= trimi <= nums[i].length

Query Kth Smallest Trimmed Number LeetCode Solution in C++

class Solution {
public:
    vector<int> smallestTrimmedNumbers(vector<string>& nums, vector<vector<int>>& queries) {
        vector<int> res;
        for(auto x:queries)
        {
            priority_queue<pair<string,int>> v;
            for(int i=0;i<nums.size();i++)
            {
                int t=nums[i].length()-x[1];
                string p=nums[i].substr(t,x[1]);
                if(v.size()<x[0])
                    v.push({p,i});
                else
                {
                    if(v.top().first > p)
                    {
                        v.pop();
                        v.push({p,i});
                    }
                }
            }
            int val=v.top().second;
            res.push_back(val);
        }
        return res;
    }
};

Query Kth Smallest Trimmed Number LeetCode Solution in Java

public int[] smallestTrimmedNumbers(String[] nums, int[][] queries) {
    HashMap<Integer, Node[]> map = new HashMap<>();
    int[] res = new int[queries.length];
    int idx = 0, len = nums[0].length();
    for(int[] query : queries){
        if(!map.containsKey(query[1])){  // make and store array if not already made.
            Node[] arr = new Node[nums.length];
            for(int i=0; i<nums.length; i++){
                String x = nums[i].substring(len-query[1], len); // trim as required
                arr[i] = new Node(i, x);
            }
			Arrays.sort(arr, (a, b)-> a.val.compareTo(b.val)); // sort array
            map.put(query[1], arr);
        }
        res[idx++] = map.get(query[1])[query[0]-1].index; // get required value.
    }
    return res;
}

class Node{ int index; String val;  // custom object to store both index and value.
    Node(int i, String v){ this.index = i; this.val = v; } }

Query Kth Smallest Trimmed Number LeetCode Solution in Python

def smallestTrimmedNumbers(A, Q): 
    m, n = len(A[0]), len(A)
    d = [list(range(n))]
    for i in range(m-1,-1,-1):
        rk = {x:j for j,x in enumerate(d[-1])}
        d.append(sorted(range(n), key=lambda x:(A[x][i],rk[x])))
    return [d[t][k-1] for k,t in Q]
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