Interleave Two Strings

easy string two pointers

Problem

Given two strings a and b, build a new string by alternately taking the next character from a and from b, beginning with a. When one of the strings runs out, append the remaining characters of the other.

Inputa = "moon", b = "sky"

Output"mskoyon"

Pairs: (m, s), (o, k), (o, y); the trailing "n" of "moon" is appended.

def interleave(a, b):
    i = j = 0
    out = []
    while i < len(a) or j < len(b):
        if i < len(a):
            out.append(a[i]); i += 1
        if j < len(b):
            out.append(b[j]); j += 1
    return "".join(out)

function interleave(a, b) {
  let i = 0, j = 0;
  let out = "";
  while (i < a.length || j < b.length) {
    if (i < a.length) out += a[i++];
    if (j < b.length) out += b[j++];
  }
  return out;
}

class Solution {
    public String interleave(String a, String b) {
        int i = 0, j = 0;
        StringBuilder out = new StringBuilder();
        while (i < a.length() || j < b.length()) {
            if (i < a.length()) out.append(a.charAt(i++));
            if (j < b.length()) out.append(b.charAt(j++));
        }
        return out.toString();
    }
}

string interleave(string a, string b) {
    int i = 0, j = 0;
    string out = "";
    while (i < (int)a.size() || j < (int)b.size()) {
        if (i < (int)a.size()) out += a[i++];
        if (j < (int)b.size()) out += b[j++];
    }
    return out;
}

Time: O(n + m) Space: O(n + m)