String is "abcded"
l =0, h = 0
i = 1, l = 0, h = 1, max = 1, A[a]=1
i = 2, l = 0, h = 2, max = 2, A[b] = 2
i = 3, l = 0, h = 3, max = 3, A[c] = 3
i = 4, l = 0, h = 4, max = 4, A[d] = 4
i = 5, l = 0, h = 5, max = 5, A[e] = 5
i = 6, 'd' is encountered again, update l = A[d] = 4, new A[d] = 5, h = 6,
max = max(5, 6-4)= max(5, 2) = 5
hence ur ans = 5
Regards
Vaibhav Mittal
Computer Science
Netaji Subhas Institute Of Technology
Delhi.
On , Interstellar Overdrive <abhi123khat...@gmail.com> wrote:
@svm11: Take the case with original string "abcded" output should be 5
but your algo will give the answer as 0.
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