Google Interview Question for Software Engineers

Country: United States

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1
of 1 vote

using binary search we can find

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0
of 0 vote

``````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)``````

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0
of 0 vote

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]``````

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-1
of 1 vote

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

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-1
of 1 vote

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))

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-1
of 1 vote

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()));
}
}``````

Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

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