Algorithm
Problem Name: 942. DI String Match
A permutation perm
of n + 1
integers of all the integers in the range [0, n]
can be represented as a string s
of length n
where:
s[i] == 'I'
ifperm[i] < perm[i + 1]
, ands[i] == 'D'
ifperm[i] > perm[i + 1]
.
Given a string s
, reconstruct the permutation perm
and return it. If there are multiple valid permutations perm, return any of them.
Example 1:
Input: s = "IDID" Output: [0,4,1,3,2]
Example 2:
Input: s = "III" Output: [0,1,2,3]
Example 3:
Input: s = "DDI" Output: [3,2,0,1]
Constraints:
1 <= s.length <= 105
s[i]
is either'I'
or'D'
.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int[] diStringMatch(String S) {
int n = S.length();
int low = 0;
int high = n;
int[] ans = new int[n + 1];
for (int i = 0; i < n; i++) {
ans[i] = S.charAt(i) == 'I' ? low++ : high--;
}
ans[n] = low;
return ans;
}
}
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#2 Code Example with Javascript Programming
Code -
Javascript Programming
const diStringMatch = function(S) {
const N = S.length
const arr = []
for(let i = 0; i <= N; i++) {
arr[i] = i
}
const res = []
for(let i = 0; i < N; i++) {
if(S[i] === 'I') {
res.push(arr.shift())
} else if(S[i] === 'D') {
res.push(arr.pop())
}
}
res.push(arr.pop())
return res
};
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#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def diStringMatch(self, S):
l, r, arr = 0, len(S), []
for s in S:
arr.append(l if s == "I" else r)
l, r = l + (s == "I"), r - (s == "D")
return arr + [l]
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#4 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0942_DIStringMatch
{
public int[] DiStringMatch(string S)
{
var result = new int[S.Length + 1];
int lo = 0, hi = S.Length;
for (int i = 0; i < S.Length; i++)
{
if (S[i] == 'I') result[i] = lo++;
else if (S[i] == 'D') result[i] = hi--;
}
result[S.Length] = lo;
return result;
}
}
}
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