Leetcode-Problems/queries-on-a-permutation-with-key.cpp at master · sajalch/Leetcode-Problems · GitHub

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/*

Given the array queries of positive integers between 1 and m, you have to process all queries[i] (from i=0 to i=queries.length-1) according to the following rules:

In the beginning, you have the permutation P=[1,2,3,...,m].
For the current i, find the position of queries[i] in the permutation P (indexing from 0) and then move this at the beginning of the permutation P. Notice that the position of queries[i] in P is the result for queries[i].
Return an array containing the result for the given queries.


Example 1:

Input: queries = [3,1,2,1], m = 5
Output: [2,1,2,1]
Explanation: The queries are processed as follow:
For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is 2, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5].
For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5].
For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is 2, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5].
For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5].
Therefore, the array containing the result is [2,1,2,1].
Example 2:

Input: queries = [4,1,2,2], m = 4
Output: [3,1,2,0]
Example 3:

Input: queries = [7,5,5,8,3], m = 8
Output: [6,5,0,7,5]


Constraints:

1 <= m <= 10^3
1 <= queries.length <= m
1 <= queries[i] <= m

*/


/*
Runtime: 24 ms, faster than 74.08% of C++ online submissions for Queries on a Permutation With Key.
Memory Usage: 8.5 MB, less than 100.00% of C++ online submissions for Queries on a Permutation With Key.

*/

class Solution {
public:
    vector<int> processQueries(vector<int>& queries, int m) {
        vector<int> p,result;
        int n=queries.size();
        bool found=false;
        for(int i=1;i<=m;i++)
        {
            p.push_back(i);
        }

        for(int i=0;i<n;i++)
        {
            int j;
            for(j=0;j<m;j++)
            {
                if(queries[i]==p[j])
                {found=true;break;}
                if(queries[i]>m||queries[i]<m-m)
                {break;}
            }
            p.insert(p.begin(),p[j]);
            p.erase(p.begin()+j+1);
            // for(int pp:p)
            // {cout<<pp<<" ";
            // }cout<<endl;
            if(found)
            {
                result.push_back(j);
            }

        }

       return result;
    }
};