Wrap every continuous instance of given word with some given string

Given two string S1 and S2. The task is to wrap every instance of string S2 in string S1 with some string on either side.

Note: Here we will use underscore(_) for the wrapping part. It can be anything as per need, for example some HTML tag, white space, etc.

Examples:

Input: S1 = “examwill be a examexam”, S2 = “exam”
Output: “_exam_will be a _examexam_”
Explanation:
String S2 is “exam“, so wrap occurrence of S2 with underscore. Since, last two occurance overlaps(in string examexam), so merge both the occurrence and wrap by putting underscore to farther left and farther right of merged substring.

Input: str = “abcabcabcabc”, substr = “abc”
Output: “_abcabcabcabc_”



Approach: The idea is to simply get the location of all instances of string S2 in S1 and add underscore at either end of the occurrence and if two or more instances of S2 overlaps then, add underscore at either end of the merged substring. Below are the steps:

  1. Get the location of all instances of the string S2 in the main string S1. For that traverse string S1 one character at a time and call substring-matching function, find().
  2. Create a 2D array of locations(say arr[][]), where each subarray holds the starting and ending indices of a specific instance of the string S2 in the string S1.
  3. Merge the overlapping starting and ending index intervals stored in arr[][].
  4. After the above steps all the overlapping indexes are merged. Now traverse the given string and intervals at the same time and create a new string by wrapping underscores.
  5. Print the new string after the above step.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function Definations
vector<vector<int> > getLocations(string str,
                                  string subStr);
  
vector<vector<int> > collapse(
    vector<vector<int> > locations);
  
string underscorify(string str,
                    vector<vector<int> > locations);
  
// Function that creates the string by
// wrapping the underscore
string underscorifySubstring(string str,
                             string subStr)
{
  
    // Function Call to intervals of
    // starting and ending index
    vector<vector<int> > locations
        = collapse(getLocations(str, subStr));
  
    // Function Call
    return underscorify(str, locations);
}
  
// Function finds all starting and ending
// index of the substring in given string
vector<vector<int> > getLocations(string str,
                                  string subStr)
{
    vector<vector<int> > locations{};
    int startIdx = 0;
  
    int L = substr.length();
  
    // Traverse string str
    while (startIdx < str.length()) {
  
        // Find substr
        int nextIdx = str.find(subStr,
                               startIdx);
  
        // If location found, then insert
        // pair int location[]
        if (nextIdx != string::npos) {
  
            locations.push_back({ nextIdx,
                                  (nextIdx + L) });
  
            // Update the start index
            startIdx = nextIdx + 1;
        }
        else {
            break;
        }
    }
    return locations;
}
  
// Function to merge the locations
// of substrings that overlap each
// other or sit next to each other
vector<vector<int> > collapse(
    vector<vector<int> > locations)
{
    if (locations.empty()) {
        return locations;
    }
  
    // 2D vector to store the merged
    // location of substrings
    vector<vector<int> > newLocations{ locations[0] };
  
    vector<int>* previous = &newLocations[0];
  
    for (int i = 1; i < locations.size(); i++) {
  
        vector<int>* current = &locations[i];
  
        // Condition to check if the
        // substring overlaps
        if (current->at(0) <= previous->at(1)) {
            previous->at(1) = current->at(1);
        }
        else {
            newLocations.push_back(*current);
            previous
                = &newLocations[newLocations.size() - 1];
        }
    }
    return newLocations;
}
  
// Function creates a new string with
// underscores added at correct positions
string underscorify(string str,
                    vector<vector<int> > locations)
{
    int locationsIdx = 0;
    int stringIdx = 0;
    bool inBetweenUnderscores = false;
    vector<string> finalChars{};
    int i = 0;
  
    // Traverse the string and check
    // in locations[] to append _
    while (stringIdx < str.length()
           && locationsIdx < locations.size()) {
  
        if (stringIdx
            == locations[locationsIdx][i]) {
  
            // Insert underscore
            finalChars.push_back("_");
            inBetweenUnderscores
                = !inBetweenUnderscores;
  
            // Increment location index
            if (!inBetweenUnderscores) {
                locationsIdx++;
            }
            i = i == 1 ? 0 : 1;
        }
  
        // Create string s
        string s(1, str[stringIdx]);
  
        // Push the created string
        finalChars.push_back(s);
        stringIdx++;
    }
  
    if (locationsIdx < locations.size()) {
        finalChars.push_back("_");
    }
    else if (stringIdx < str.length()) {
        finalChars.push_back(str.substr(stringIdx));
    }
  
    // Return the resultant string
    return accumulate(finalChars.begin(),
                      finalChars.end(), string());
}
  
// Driver Code
int main()
{
    // Given string S1 and S2
    string S1 = "examwill be a examexam";
    string S2 = "exam";
  
    // Function Call
    cout << underscorifySubstring(S1, S2);
    return 0;
}

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Output:

_exam_will be a _examexam_


Time Complexity: O(N * M), where N and M is the length of the string S1 and S2 respectively.
Auxiliary Space: O(N), where N is the length of string S1

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