Approach: Count magazine letters

def canConstruct(s, t):
    from collections import Counter
    have = Counter(t)
    for c in s:
        if have[c] <= 0:
            return False
        have[c] -= 1
    return True

Complexity: O(s + t) time, O(1) space.

Full Solution — all languages

class Solution:
    def canConstruct(self, s, t):
        have = {}
        for c in t:
            have[c] = have.get(c, 0) + 1
        for c in s:
            if have.get(c, 0) <= 0:
                return False
            have[c] -= 1
        return True

import java.util.*;
class Solution {
    public boolean canConstruct(String s, String t) {
        int[] have = new int[128];
        for (char c : t.toCharArray()) have[c]++;
        for (char c : s.toCharArray()) {
            if (have[c] <= 0) return false;
            have[c]--;
        }
        return true;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: bool canConstruct(string s, string t) {
        int have[128] = {0};
        for (char c : t) have[(int)c]++;
        for (char c : s) {
            if (have[(int)c] <= 0) return false;
            have[(int)c]--;
        }
        return true;
    }
};

/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */ var canConstruct = function(s, t) {
    const have = {};
    for (const c of t) have[c] = (have[c] || 0) + 1;
    for (const c of s) {
        if (!have[c]) return false;
        have[c]--;
    }
    return true;
};

Ransom Note

Example 1

Example 2

Constraints