## Longest Common Prefix Difficulty: Hard

#### Understanding the problem

Problem Description

Given the array of strings S, write a program to find the longest common prefix string which is the prefix of all the strings in the array.

Problem Note

• The longest common prefix for a pair of strings S1 and S2 is the longest string which is the prefix of both S1 and S2.
• All given inputs are in lowercase letters a-z.
• If there is no common prefix, return `"-1"`.

Example 1

``````Input: S[] = [“apple", "ape", "april”]
Output: "ap"``````

Example 2

``````Input: S[] = ["flower","flow","flight"]
Output: "fl"``````

Example 3

``````Input: S[] = [“after”, ”academy, ”mindorks”]
Output: “-1”
Explanation: There is no common prefix among the input strings.``````

#### Solutions

We will be discussing four different approaches to solve this problem

1. Horizontal Scanning — Find the LCP of strs with strs with strs and so on.
2. Vertical Scanning — Scan all the characters at index 0 then index 1 then index 2 and so on.
3. Divide and Conquer — Divide the strs array to two parts and merge the LCP of both the subparts.
4. Binary Search — Compare the substrings[0 to mid] of the smallest string with each string and keep on updating `front` and `behind` accordingly.

#### 1. Horizontal Scanning

A simple way to find the longest common prefix shared by a set of strings LCP(S1​…Sn​) could be found under the observation that
`LCP(S1​…Sn​) = LCP(LCP(LCP(S1​, S2​), S3​), ….Sn​)`

To achieve it, simply iterate through the strings [S1​…Sn​], finding at each iteration `i` the longest common prefix of strings LCP(S1​…Si​). When the LCP(S1​…Si​) is an empty string, then you can return an empty string. Otherwise, after n iterations, the algorithm will returns LCP(S1​…Sn​).

Solution steps
• take a variable `prefix` and initialize it with `strs`
• compare the prefix of each string in the `strs` array with the `prefix` variable and update the `prefix` accordingly
• if at any point length of `prefix` becomes `0` then return `-1`
• return `prefix` after comparing with each string of the `strs` array
Pseudo Code
``````string longestCommonPrefix(string[] strs, int size) {
if (size == 0)
return "-1"
string prefix = strs
for (int i = 1 to size)
while (strs[i].indexOf(prefix) != 0) {
prefix = prefix.substring(0, prefix.length() - 1)
if (prefix.isEmpty())
return "-1"
}
return prefix
}``````
Complexity Analysis

Time complexity: O(S), where S is the sum of all characters in all strings.

Space complexity: O(1).

Critical Ideas to Think
• Do you think that if all the strings in the array would be same then it would be the worst-case for this approach? If yes, Why?
• What does the `prefix.substring(0, prefix.length() -1)` mean?
• What is the initial value of `prefix` the variable and why?
• Can you think of any other approach?

#### 2. Vertical scanning

Imagine a very short string is the common prefix of the array. The above approach will still do S comparisons. One way to optimize this case is to do vertical scanning. We compare characters from top to bottom on the same column (same character index of the strings) before moving on to the next column.

Example —

``````[
"AfterLife",
"Affirmative",
]``````
``Vertically scan the 0th index of all the strings, in this case its "A" for each string so the LCP till now is "A". Now vertically scan the 1st index of each string, the second character for each string are not same, so we will return the prefix till now.``
Solution Step

Start comparing the `ith` character for each string, if all the character for `ith` position are all same, then add it to the prefix, otherwise, return prefix till now.

Pseudo Code
``````string longestCommonPrefix(String[] strs, int size) {
if (strs.length == 0)
return "-1"
for (int i = 0 to i < strs.length()){
char c = strs[i]
for (int j = 1 to j < strs.length) {
if (i == strs[j].length() or strs[j][i] != c)
return strs.substring(0, i)
}
}
return strs
}``````
Complexity Analysis

Time complexity: O(S), where S is the sum of all characters in all strings.

In the best case there are at most `n*minLen` comparisons where `minLen` is the length of the shortest string in the array.

Space complexity: O(1). We only used constant extra space.

Critical Ideas to Think
• Do you think the worst case for this approach is exactly the same as in the horizontal scanning?
• Why we are returning `strs` at the end of the function in pseudocode?

#### 3: Divide and conquer

The thought of this algorithm is related to the associative property of LCP operation. Notice that: `LCP(S1​…Sn​) = LCP(LCP(S1​…Sk​), LCP(Sk+1​…Sn​))`, where `LCP(S1​…Sn​)` is the longest common prefix in the set of strings `[S1​…Sn​]`, `1 < k < n1 < k < n`

Thus, the divide and conquer approach could be implied here by dividing the `LCP(Si…Sj)` problem into two subproblems `LCP(Si​…Smid​)` and `LCP(Smid+1​…Sj​)`, where `mid` is the middle of the `Si` and `Sj` .

We can keep on dividing the problems into two subproblems until they cannot be divided further.

Now to conquer the solution, we compare the solutions of the two subproblems till there is no character match at each level. The found common prefix would be the solution of `LCP(Si…Sj)`. Solution Steps
• Recursively divide the strs array into two sub-arrays.
• In the conquer step, merge the result of the two sub-arrays which will be `LCP(LCP(strs[left…mid], LCP([mid+1…right]))` and return it.
Pseudo Code
``````string longestCommonPrefix(string[] strs, int size) {
if (size == 0) return "-1"
return longestCommonPrefixutil(strs, 0 , size - 1)
}``````
``````string longestCommonPrefixutil(string[] strs, int left, int right) {
if (left == right) {
return strs[left]
}
else {
int mid = (left + right)/2;
string left_lcp =   longestCommonPrefixutil(strs, left , mid)
string right_lcp =  longestCommonPrefixutil(strs, mid + 1,right)
return commonPrefix(left_lcp, right_lcp)
}
}``````
``````string commonPrefix(String left, String right) {
int smaller = min(left.length(), right.length())
for (int i = 0 to i < smaller) {
if ( left[i] != right[i] )
return left.substring(0, i)
}
return left.substring(0, smaller)
}``````
Complexity Analysis

Time complexity: O(S), where S is the number of all characters in the array.

Space complexity: O(m ⋅ logn) (Why?)

There are log n recursive calls and each store need m space to store the result

Critical ideas to think
• How we are dividing the problems set to subproblems?
• Do you think divide and conquer is similar to horizontal scanning?
• Do you think that the best case complexity will be `O(minLen*n)` ?
• What does the `commonPrefix `function do?

#### 4: Binary search

The idea is to apply a binary search method to find the string with maximum value `L`, which is the common prefix of all of the strings. The algorithm searches space is the interval (0…minLen), where `minLen` is minimum string length and the maximum possible common prefix. Each time the search space is divided into two equal parts, one of them is discarded because it is sure that it doesn't contain the solution. There are two possible cases: `S[1...mid]` is not a common string. This means that for each `j > i, S[1..j]` is not a common string and we discard the second half of the search space. `S[1...mid]` is a common string. This means that for each `i < j, S[1..i]` is a common string and we discard the first half of the search space because we try to find a longer common prefix.

Solution steps
1. Pick the smallest string
2. Take variable `front = 0 `and` behind = len(smallest string)`
3. Do binary search
• Compare the substring up to middle character of the smallest string with every other string at that index.
• if all the strings have the same substring(0, mid) then move `front` to `mid + 1` else move `behind` to `mid-1` .
• If `front `becomes equal to `behind `then return the `strs.substring(0, mid)`
Pseudo Code
``````string longestCommonPrefix(string[] strs, int size) {
if(size==0)
return "-1"
minLen=INT_MAX
for(i = 0 to i < size){
minLen = min(minLen,strs[i].length())
}
front = 1
behind = minLen``````
``````    prefix = ""
while(front <= behind) {
mid = (front+behind)/2
string temp=strs.substring(0, mid)
int j = 1
for(j=1 to j < size) {
if(!strs[j].startsWith(temp)) {
behind = mid-1
break
}
}
if(j==strs.length){``````
``````            prefix = temp
front=mid+1
}
}
return prefix
}``````
Complexity Analysis

Time complexity: O(S⋅logn), where S is the sum of all characters in all strings

Space complexity: O(1)

Critical Ideas to Think
• Do you think that the binary search approach is not better than the approaches described above?
• Why we are comparing substrings(0 to mid) instead of comparing only the middle character of every other string in the strs array? Can you think of a case in this scenario when we will compare only the mid character?
• Why did we start this algorithm by finding the minLen?
• Do you think that the best case and average case are the same in the binary search approach?
• Can you take some example and compare the time complexity of each of the approaches described above.

#### Comparison of Different Approaches #### Suggested Problems to Solve

• Longest Common Prefix using Trie
• Find the shortest unique prefix for every word in the given list
• Find Longest common prefix using linked list
• Find minimum shift for longest common prefix

If you have any more approaches or you find an error/bug in the above solutions, please comment down below.

Happy Coding! Enjoy Algorithms!