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CustomSortString.java
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CustomSortString.java
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/* (C) 2024 YourCompanyName */
package hashing;
import java.util.*;
/**
* Created by gouthamvidyapradhan on 01/05/2018. S and T are strings composed of lowercase letters.
* In S, no letter occurs more than once.
*
* <p>S was sorted in some custom order previously. We want to permute the characters of T so that
* they match the order that S was sorted. More specifically, if x occurs before y in S, then x
* should occur before y in the returned string.
*
* <p>Return any permutation of T (as a string) that satisfies this property.
*
* <p>Example : Input: S = "cba" T = "abcd" Output: "cbad" Explanation: "a", "b", "c" appear in S,
* so the order of "a", "b", "c" should be "c", "b", and "a". Since "d" does not appear in S, it can
* be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.
*
* <p>Note:
*
* <p>S has length at most 26, and no character is repeated in S. T has length at most 200. S and T
* consist of lowercase letters only.
*
* <p>Solution: O(N) count occurrence of each character and write to the output string
*/
public class CustomSortString {
/**
* Main method
*
* @param args
* @throws Exception
*/
public static void main(String[] args) throws Exception {
System.out.println(new CustomSortString().customSortString("cba", "abcd"));
}
public String customSortString(String S, String T) {
Map<Character, Integer> map = new HashMap<>();
for (int i = 0; i < T.length(); i++) {
if (!map.containsKey(T.charAt(i))) {
map.put(T.charAt(i), 1);
} else {
map.put(T.charAt(i), map.get(T.charAt(i)) + 1);
}
}
StringBuilder result = new StringBuilder();
for (char c : S.toCharArray()) {
if (map.containsKey(c)) {
int count = map.remove(c);
for (int i = 0; i < count; i++) {
result.append(c);
}
}
}
for (char c : map.keySet()) {
int count = map.get(c);
for (int i = 0; i < count; i++) {
result.append(c);
}
}
return result.toString();
}
}