The Custom Sort String Algorithm with Count and Write

S and T are strings composed of lowercase letters. In S, no letter occurs more than once. 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.

Return any permutation of T (as a string) that satisfies this property.
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.

Note:

• 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.

Count and Write Algorithm

The S has the orders of the characters that appear in the string T thus if we go through the S character by character, and only print those that appear in string T, the order can be preserved. In order to do this, we need to count the letters in T.

When the first step is done, we need to print those letters that do not appear in string S. This is done by going through string T and check if has appeared in S. Since the input S and T are both lowercase letters, we only require 2 static size array to do the counting.

The space complexity is O(1) and the time complexity is O(N). If the input string can be more than lowercase letters, we might use a hash table to do the counting, in which case the space complexity is O(N).

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class Solution {
public:
    string customSortString(string S, string T) {
        int s[26] = { 0 };
        int t[26] = { 0 };
        for (const auto n: S) {
            s[n - 'a'] ++;
        }
        for (const auto n: T) {
            t[n - 'a'] ++;
        }
        string r = "";
        for (int i = 0; i < S.size(); ++ i) {
            for (int k = 0; k < t[S[i] - 'a']; ++ k) {
                r += S[i];
            }
        }
        for (int i = 0; i < T.size(); ++ i) {
            if (s[T[i] - 'a'] == 0) {
                r += T[i];
            }
        }
        return r;
    }
};

class Solution {
public:
    string customSortString(string S, string T) {
        int s[26] = { 0 };
        int t[26] = { 0 };
        for (const auto n: S) {
            s[n - 'a'] ++;
        }
        for (const auto n: T) {
            t[n - 'a'] ++;
        }
        string r = "";
        for (int i = 0; i < S.size(); ++ i) {
            for (int k = 0; k < t[S[i] - 'a']; ++ k) {
                r += S[i];
            }
        }
        for (int i = 0; i < T.size(); ++ i) {
            if (s[T[i] - 'a'] == 0) {
                r += T[i];
            }
        }
        return r;
    }
};

We can reduce the above to using only 1 array of 26, by iterating from ‘a’ to ‘z’ and print those who do not appear in S (that are printed already should not be printed again).

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class Solution {
public:
    string customSortString(string S, string T) {
        int t[26] = { 0 };
        for (const auto n: T) {
            t[n - 'a'] ++;
        }
        string r = "";
        for (int i = 0; i < S.size(); ++ i) {
            for (int k = 0; k < t[S[i] - 'a']; ++ k) {
                r += S[i];
            }          
            t[S[i] - 'a'] = 0; // mark those used
        }
        for (char c = 'a'; c <= 'z'; ++ c) {
            for (int k = 0; k < t[c - 'a']; ++ k) {
                r += c;
            }
        }
        return r;
    }
};

class Solution {
public:
    string customSortString(string S, string T) {
        int t[26] = { 0 };
        for (const auto n: T) {
            t[n - 'a'] ++;
        }
        string r = "";
        for (int i = 0; i < S.size(); ++ i) {
            for (int k = 0; k < t[S[i] - 'a']; ++ k) {
                r += S[i];
            }          
            t[S[i] - 'a'] = 0; // mark those used
        }
        for (char c = 'a'; c <= 'z'; ++ c) {
            for (int k = 0; k < t[c - 'a']; ++ k) {
                r += c;
            }
        }
        return r;
    }
};

–EOF (The Ultimate Computing & Technology Blog) —