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Let's now go ahead and discuss the pseudocode for the recursive approach that we have just now discussed

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in the previous video.

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So we would need to write a function.

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And this function will have some inputs.

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And inside this function we would have to write the base case and the recursive case.

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Now we have already discussed the three possibilities for the base case.

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Now when it comes to these inputs over here we'll need two indices index one, which can be used as

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a pointer to iterate over the characters in word one.

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And let's say we are using index two to iterate over the characters in word two.

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Now in the base case, we have three possibilities.

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So if both of these become greater than the last valid indices in word one and word two together, then

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we can just return zero, or if only one of them is greater than the last valid index, and there are

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still valid characters in the other word, then we would have to return the number of remaining characters.

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In the other word.

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So we have discussed this in the previous video.

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And when it comes to the choice diagram, we have discussed that there are two possibilities at a high

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level.

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One is that we find that the characters at index one and index two in word one and word two, respectively,

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are equal.

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And the other possibility is that we see that the two characters are not equal.

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Now, if they are equal, we don't need to do any operations, and we can just find the number of operations

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required for the remaining parts of word one and word two, respectively.

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Now, if they are not equal, we have discussed that there are three possibilities insert, delete and

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replace.

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Now we have seen how to move the indices index one and index two in these three cases in the previous

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video.

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So for each of these we would have to find the answer.

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So insert would be one plus.

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Again recursively calling the function with the respective indices.

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As we have discussed in the previous video.

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And we would have to do the same thing for delete and replace.

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And then we would have to return the minimum among these three values that we get.

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So this is how we will recursively solve this question.

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And this is the pseudocode for the approach that we have discussed.