Knuth–Morris–Pratt algorithm

To illustrate the algorithm's details, consider a (relatively artificial) run of the algorithm, where W = "ABCDABD" and S = "ABC ABCDAB ABCDABCDABDE". At any given time, the algorithm is in a state determined by two integers:

• m, denoting the position within S where the prospective match for W begins,
• i, denoting the index of the currently considered character in W.

In each step the algorithm compares S[m+i] with W[i] and increments i if they are equal. This is depicted, at the start of the run, like

             1         2  
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W: ABCDABD
i: 0123456

The algorithm compares successive characters of W to "parallel" characters of S, moving from one to the next by incrementing i if they match. However, in the fourth step S[3] = ' ' does not match W[3] = 'D'. Rather than beginning to search again at S[1], we note that no 'A' occurs between positions 1 and 2 in S; hence, having checked all those characters previously (and knowing they matched the corresponding characters in W), there is no chance of finding the beginning of a match. Therefore, the algorithm sets m = 3 and i = 0.

             1         2  
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:    ABCDABD
i:    0123456

This match fails at the initial character, so the algorithm sets m = 4 and i = 0

             1         2  
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:     ABCDABD
i:     0123456

Here, i increments through a nearly complete match "ABCDAB" until i = 6 giving a mismatch at W[6] and S[10]. However, just prior to the end of the current partial match, there was that substring "AB" that could be the beginning of a new match, so the algorithm must take this into consideration. As these characters match the two characters prior to the current position, those characters need not be checked again; the algorithm sets m = 8 (the start of the initial prefix) and i = 2 (signaling the first two characters match) and continues matching. Thus the algorithm not only omits previously matched characters of S (the "AB"), but also previously matched characters of W (the prefix "AB").

             1         2  
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:         ABCDABD
i:         0123456

This search at the new position fails immediately because W[2] (a 'C') does not match S[10] (a ' '). As in the first trial, the mismatch causes the algorithm to return to the beginning of W and begins searching at the mismatched character position of S: m = 10, reset i = 0.

             1         2  
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:           ABCDABD
i:           0123456

The match at m=10 fails immediately, so the algorithm next tries m = 11 and i = 0.

             1         2  
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:            ABCDABD
i:            0123456

Once again, the algorithm matches "ABCDAB", but the next character, 'C', does not match the final character 'D' of the word W. Reasoning as before, the algorithm sets m = 15, to start at the two-character string "AB" leading up to the current position, set i = 2, and continue matching from the current position.

             1         2  
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:                ABCDABD
i:                0123456

This time the match is complete, and the first character of the match is S[15].

The above example contains all the elements of the algorithm. For the moment, we assume the existence of a "partial match" table T, described below, which indicates where we need to look for the start of a new match when a mismatch is found. The entries of T are constructed so that if we have a match starting at S[m] that fails when comparing S[m + i] to W[i], then the next possible match will start at index m + i - T[i] in S (that is, T[i] is the amount of "backtracking" we need to do after a mismatch). This has two implications: first, T[0] = -1, which indicates that if W[0] is a mismatch, we cannot backtrack and must simply check the next character; and second, although the next possible match will begin at index m + i - T[i], as in the example above, we need not actually check any of the T[i] characters after that, so that we continue searching from W[T[i]]. The following is a sample pseudocode implementation of the KMP search algorithm.

algorithm kmp_search:
    input:
        an array of characters, S (the text to be searched)
        an array of characters, W (the word sought)
    output:
        an array of integers, P (positions in S at which W is found)
        an integer, nP (number of positions)

    define variables:
        an integer, j ← 0 (the position of the current character in S)
        an integer, k ← 0 (the position of the current character in W)
        an array of integers, T (the table, computed elsewhere)

    let nP ← 0

    while j < length(S) do
        if W[k] = S[j] then
            let j ← j + 1
            let k ← k + 1
            if k = length(W) then
                (occurrence found, if only first occurrence is needed, m ← j - k  may be returned here)
                let P[nP] ← j - k, nP ← nP + 1
                let k ← T[k] (T[length(W)] can't be -1)
        else
            let k ← T[k]
            if k < 0 then
                let j ← j + 1
                let k ← k + 1

Assuming the prior existence of the table T, the search portion of the Knuth–Morris–Pratt algorithm has complexity O(n), where n is the length of S and the O is big-O notation. Except for the fixed overhead incurred in entering and exiting the function, all the computations are performed in the while loop. To bound the number of iterations of this loop; observe that T is constructed so that if a match which had begun at S[m] fails while comparing S[m + i] to W[i], then the next possible match must begin at S[m + (i - T[i])]. In particular, the next possible match must occur at a higher index than m, so that T[i] < i.

This fact implies that the loop can execute at most 2n times, since at each iteration it executes one of the two branches in the loop. The first branch invariably increases i and does not change m, so that the index m + i of the currently scrutinized character of S is increased. The second branch adds i - T[i] to m, and as we have seen, this is always a positive number. Thus the location m of the beginning of the current potential match is increased. At the same time, the second branch leaves m + i unchanged, for m gets i - T[i] added to it, and immediately after T[i] gets assigned as the new value of i, hence new_m + new_i = old_m + old_i - T[old_i] + T[old_i] = old_m + old_i. Now, the loop ends if m + i = n; therefore, each branch of the loop can be reached at most n times, since they respectively increase either m + i or m, and m ≤ m + i: if m = n, then certainly m + i ≥ n, so that since it increases by unit increments at most, we must have had m + i = n at some point in the past, and therefore either way we would be done.

Thus the loop executes at most 2n times, showing that the time complexity of the search algorithm is O(n).

Here is another way to think about the runtime: Let us say we begin to match W and S at position i and p. If W exists as a substring of S at p, then W[0..m] = S[p..p+m]. Upon success, that is, the word and the text matched at the positions (W[i] = S[p+i]), we increase i by 1. Upon failure, that is, the word and the text do not match at the positions (W[i] ≠ S[p+i]), the text pointer is kept still, while the word pointer is rolled back a certain amount (i = T[i], where T is the jump table), and we attempt to match W[T[i]] with S[p+i]. The maximum number of roll-back of i is bounded by i, that is to say, for any failure, we can only roll back as much as we have progressed up to the failure. Then it is clear the runtime is 2n.

We consider the example of W = "ABCDABD" first. We will see that it follows much the same pattern as the main search, and is efficient for similar reasons. We set T[0] = -1. To find T[1], we must discover a proper suffix of "A" which is also a prefix of pattern W. But there are no proper suffixes of "A", so we set T[1] = 0. To find T[2], we see that the substring W[0] - W[1] ("AB") has a proper suffix "B". However "B" is not a prefix of the pattern W. Therefore, we set T[2] = 0.

Continuing to T[3], we first check the proper suffix of length 1, and as in the previous case it fails. Should we also check longer suffixes? No, we now note that there is a shortcut to checking all suffixes: let us say that we discovered a proper suffix which is a proper prefix (A proper prefix of a string is not equal to the string itself) and ending at W[2] with length 2 (the maximum possible); then its first character is also a proper prefix of W, hence a proper prefix itself, and it ends at W[1], which we already determined did not occur as T[2] = 0 and not T[2] = 1. Hence at each stage, the shortcut rule is that one needs to consider checking suffixes of a given size m+1 only if a valid suffix of size m was found at the previous stage (i.e. T[x] = m) and should not bother to check m+2, m+3, etc.

Therefore, we need not even concern ourselves with substrings having length 2, and as in the previous case the sole one with length 1 fails, so T[3] = 0.

We pass to the subsequent W[4], 'A'. The same logic shows that the longest substring we need to consider has length 1, and as in the previous case it fails since "D" is not a prefix of W. But instead of setting T[4] = 0, we can do better by noting that W[4] = W[0], and also that a look-up of T[4] implies the corresponding S character, S[m+4], was a mismatch and therefore S[m+4] ≠ 'A'. Thus there is no point in restarting the search at S[m+4]; we should begin at 1 position ahead. This means that we may shift pattern W by match length plus one character, so T[4] = -1.

Considering now the next character, W[5], which is 'B': though by inspection the longest substring would appear to be 'A', we still set T[5] = 0. The reasoning is similar to why T[4] = -1. W[5] itself extends the prefix match begun with W[4], and we can assume that the corresponding character in S, S[m+5] ≠ 'B'. So backtracking before W[5] is pointless, but S[m+5] may be 'A', hence T[5] = 0.

Finally, we see that the next character in the ongoing segment starting at W[4] = 'A' would be 'B', and indeed this is also W[5]. Furthermore, the same argument as above shows that we need not look before W[4] to find a segment for W[6], so that this is it, and we take T[6] = 2.

Therefore, we compile the following table:

Another example:

Another example (slightly changed from the previous example):

Another more complicated example:

The example above illustrates the general technique for assembling the table with a minimum of fuss. The principle is that of the overall search: most of the work was already done in getting to the current position, so very little needs to be done in leaving it. The only minor complication is that the logic which is correct late in the string erroneously gives non-proper substrings at the beginning. This necessitates some initialization code.

algorithm kmp_table:
    input:
        an array of characters, W (the word to be analyzed)
    output:
        an array of integers, T (the table to be filled)

    define variables:
        an integer, pos ← 1 (the current position we are computing in T)
        an integer, cnd ← 0 (the zero-based index in W of the next character of the current candidate substring)

    let T[0] ← -1

    while pos < length(W) do
        if W[pos] = W[cnd] then
            let T[pos] ← T[cnd]
        else
            let T[pos] ← cnd
            while cnd ≥ 0 and W[pos] ≠ W[cnd] do
                let cnd ← T[cnd]
        let pos ← pos + 1, cnd ← cnd + 1

    let T[pos] ← cnd (only needed when all word occurrences are searched)

The time (and space) complexity of the table algorithm is ${\displaystyle O(k)}$, where ${\displaystyle k}$ is the length of W.

• The outer loop: pos is initialized to 1, the loop condition is pos < k, and pos is increased by 1 in every iteration of the loop. Thus the loop will take ${\displaystyle k-1}$ iterations.

• The inner loop: cnd is initialized to 0 and gets increased by at most 1 in each outer loop iteration. T[cnd] is always less than cnd, so cnd gets decreased by at least 1 in each inner loop iteration; the inner loop condition is cnd ≥ 0. This means that the inner loop can execute at most as many times in total, as the outer loop has executed – each decrease of cnd by 1 in the inner loop needs to have a corresponding increase by 1 in the outer loop. Since the outer loop takes ${\displaystyle k-1}$ iterations, the inner loop can take no more than ${\displaystyle k-1}$ iterations in total.

Combined, the outer and inner loops take at most ${\displaystyle 2k-2}$ iterations. This corresponds to ${\displaystyle O(k)}$ time complexity using the Big O notation.