CareerCup

using binary search we can find

def diff_char(s1, s2, idx=0):
	if idx >= len(s1) or idx >= len(s2) or s1[idx] != s2[idx]:
		return "MISMATCH AT INDEX {}".format(idx)
	return diff_char(s1, s2, idx+1)

Time complexity O(min(n, m))

def extra_char(s1: str, s2:str) -> str:
    max = s1
    min = s2
    if len(s2) > len(max):
        max = s2
        min = s1
    for c1, c2 in zip(max, min):  # O(min(n, m))
        if c1 != c2:
            return c1
    return max[-1]

we can xor both strings individual ascii code . the result will the ascii code of the extra character

def findExtraChar(s,t):
if s in t:
return t.replace(s,'')
if t in s:
return s.replace(t,'')

"""xor two strings together"""
res = 0
# end and xor with res
for i in range(len(s)) :
# xor with res
res =res ^ (ord)(s[i])
# end and xor with res
for i in range(len(t)) :
# xor with res
res = res ^ (ord)(t[i])
return ((chr)(res))

If the extra character is always at end then....

public static void main(String[] args) {

		String A = "abcjdfdkjsf";

		String B = "abcjdfdkjsfg";
		
		if(A.contains(B)) {
			System.out.println(A.substring(B.length()));
		}else if(B.contains(A)) {
			System.out.println(B.substring(A.length()));
		}
	}