Longest Common Prefix in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the first string as the initial prefix.

DSA C

char* longestCommonPrefix(char** strs, int strsSize) {
    if (strsSize == 0) return "";
    char* prefix = [1];
    return prefix;
}

Drag options to blanks, or click blank then click option'

ANULL

Bstrs[0]

C""

Dstrs[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using strs[1] instead of strs[0]

Returning NULL instead of a string

Returning an empty string "" immediately

The longest common prefix starts as the first string in the array.

2fill in blank

medium

Complete the code to loop through each string in the array.

DSA C

for (int i = [1]; i < strsSize; i++) {
    // compare prefix with strs[i]
}

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A1

B0

CstrsSize

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop at 0 causing redundant comparison

Starting loop at strsSize causing no iteration

Starting loop at -1 causing invalid index

Start from index 1 because prefix is initialized with strs[0].

3fill in blank

hard

Fix the error in the while loop condition to compare characters correctly.

DSA C

while (prefix[j] && strs[i][j] && prefix[j] == [1]) {
    j++;
}

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Astrs[i][j]

B'\0'

Cstrs[0][j]

Dprefix[j]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing prefix[j] with itself

Comparing with strs[0][j] which is prefix itself

Comparing with '\0' which is end of string

We compare prefix[j] with strs[i][j] to find common characters.

4fill in blank

hard

Fill both blanks to update prefix length and terminate it correctly.

DSA C

prefix[[1]] = '\0';
int j = [2];

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Aj

B0

C1

DstrsSize

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of j to terminate prefix

Not resetting j to 0 causing wrong indexing

Using strsSize as index causing out of bounds

Set prefix[j] to '\0' to shorten prefix, and reset j to 0 for next iteration.

5fill in blank

hard

Fill all three blanks to complete the longestCommonPrefix function.

DSA C

char* longestCommonPrefix(char** strs, int strsSize) {
    if (strsSize == 0) return "";
    char* prefix = [1];
    int i = [2];
    int j = 0;
    while (i < strsSize) {
        while (prefix[j] && strs[i][j] && prefix[j] == strs[i][j]) {
            j++;
        }
        prefix[[3]] = '\0';
        i++;
        j = 0;
    }
    return prefix;
}

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Astrs[0]

B1

Cj

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting i at 0 causing redundant comparison

Terminating prefix at 0 or wrong index

Not initializing prefix correctly

Initialize prefix with strs[0], start i at 1, and terminate prefix at j.