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### by Sumit Chourasia | Oct 01, 2020 | Category :coding | Tags : algorithmडेटा-संरचनाआसानleetcodeगणित

#### DI String Match - Math - Easy - LeetCode - मिनी टीवी

Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length.

Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1:

If S[i] == "I", then A[i] < A[i+1]
If S[i] == "D", then A[i] > A[i+1]

Example 1:

Input: "IDID"
Output: [0,4,1,3,2]
Example 2:

Input: "III"
Output: [0,1,2,3]
Example 3:

Input: "DDI"
Output: [3,2,0,1]

Note:

1 <= S.length <= 10000
S only contains characters "I" or "D".

``````public class Solution {
public int[] DiStringMatch(string S) {
int left = 0;
int right = S.Length;
var result = new int[S.Length+1];
for(int i=0;i<S.Length;i++){
if(S[i]=='I'){
result[i]=left++;
}
else{
result[i]=right--;
}
}
result[S.Length]=left;
return result;
}
}``````

Time Complexity: O(n)

Space Complexity: O(1)

Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia