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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:
nums
to its rightmost trimi
digits.kith
smallest trimmed number in nums
. If two trimmed numbers are equal, the number with the lower index is considered to be smaller.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:
x
digits means to keep removing the leftmost digit, until only x
digits remain.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.nums[i].length
are equal.1 <= queries.length <= 100
queries[i].length == 2
1 <= ki <= nums.length
1 <= trimi <= nums[i].length
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;
}
};
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; } }
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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