String Algorithms for the day before your Coding Interview (Day before coding Interview Book 4) by Kiao Ue & Chatterjee Aditya

Author:Kiao, Ue & Chatterjee, Aditya [Kiao, Ue]
Language: eng
Format: azw3
Publisher: OpenGenus
Published: 2020-05-09T16:00:00+00:00

for i from 0 to M + 1

for j from 0 to N + 1

if (i == 0 or j == 0 )

dp[i][j] = 0

else if (S1[i - 1 ] == S2[j - 1 ])

dp[i][j] = dp[i - 1 ][j - 1 ] + 1

if (result < dp[i][j])

{

result = dp[i][j];

r = i;

c = j;

}

else

dp[i][j] = 0

result is our answer

The basic condition of calculating dp[][] is result of current row in matrix dp[][] depends on values from previous row. Hence the required length of longest common substring can be obtained by maintaining values of two consecutive rows only , thereby reducing space requirements to O(2 * N) .

This would result in the following pseudocode:

dp[][] = 2D array 2 x M

result = 0 , r = 0 , c = 0

current = 0

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