Generate string after adding spaces at specific positions in a given String

Last Updated : 24 Feb, 2023

Given a string s and an array spaces[] describing the original string’s indices where spaces will be added. The task is to add spaces in given positions in spaces[] and print the string formed.

Examples:

Input: s = “GeeksForGeeK”, spaces = {1, 5, 10}
Output: “G eeks ForGe eK”
Explanation: The underlined characters in “GeeksForGeeK” relate to the indices 1, 5, and 10. After that, put spaces in front of those characters.

Input: s = “ilovegeeksforgeek”, spaces = {1, 5, 10, 13}
Output: “i love geeks for geek”

Approach#1: This problem is simple string implementation based. Follow the steps below to solve the given problem.

Initialize with a space in the new string of size of the sum of the length of both arrays.
Visit the index and wherever the index is equal to the current space value in the space[] array skip it as space is already there.
Else keep assigning the value from the main string
Here addition of ‘l’ in line { if(l<N and i==sp[l]+l) } is very interesting to observe:
- The values in the space array are basically according to the old input string.
- But in the new string, these space values or indices are basically shifting by the number of spaces found before.
Print the string formed at the end.

Below is the implementation of the above approach

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to add space in required positions
string spaceintegration(string s, vector<int>& sp)
{
    int M = s.size(), N = sp.size(), l = 0, r = 0;
 
    string res(M + N, ' ');
 
    // Iterate over M+N length
    for (int i = 0; i < M + N; i++) {
 
        if (l < N and i == sp[l] + l)
            l++;
        else
            res[i] = s[r++];
    }
 
    // Return the required string
    return res;
}
 
// Driver Code
int main()
{
 
    string s = "ilovegeeksforgeeks";
 
    vector<int> space = { 1, 5, 10, 13 };
 
    // Function Call
    cout << spaceintegration(s, space) << endl;
 
    return 0;
}

Java

// Java program for above approach
import java.util.*;
class GFG
{
 
  // Function to add space in required positions
  static String spaceintegration(String s, int []sp)
  {
    int M = s.length(), N = sp.length, l = 0, r = 0;
    String res = newstr(M + N, ' ');
 
    // Iterate over M+N length
    for (int i = 0; i < M + N; i++) {
 
      if (l < N && i == sp[l] + l)
        l++;
      else
        res = res.substring(0,i)+s.charAt(r++)+res.substring(i+1);
    }
 
    // Return the required String
    return res;
  }
 
  static String newstr(int i, char c) {
    String str = "";
    for (int j = 0; j < i; j++) {
      str+=c;        
    }
    return str;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    String s = "ilovegeeksforgeeks";
    int[] space = { 1, 5, 10, 13 };
 
    // Function Call
    System.out.print(spaceintegration(s, space) +"\n");
 
  }
}
 
// This code contributed by shikhasingrajput

Python3

# Python3 program for above approach
 
# Function to add space in required positions
def spaceintegration(s, sp):
     
    M = len(s)
    N = len(sp)
    l = 0
    r = 0
 
    res = [' '] * (M + N)
 
    # Iterate over M+N length
    for i in range(M + N):
        if (l < N and i == sp[l] + l):
            l += 1
        else:
            res[i] = s[r]
            r += 1
 
    # Return the required string
    return ''.join(res)
 
# Driver Code
if __name__ == "__main__":
 
    s = "ilovegeeksforgeeks"
 
    space = [ 1, 5, 10, 13 ]
 
    # Function Call
    print(spaceintegration(s, space))
 
# This code is contributed by ukasp

C#

// C# program for above approach
using System;
class GFG
{
 
  // Function to add space in required positions
  static String spaceintegration(String s, int []sp)
  {
    int M = s.Length, N = sp.Length, l = 0, r = 0;
    String res = newstr(M + N, ' ');
 
    // Iterate over M+N length
    for (int i = 0; i < M + N; i++) {
 
      if (l < N && i == sp[l] + l)
        l++;
      else
        res = res.Substring(0,i)+s[r++]+res.Substring(i+1);
    }
 
    // Return the required String
    return res;
  }
 
  static String newstr(int i, char c) {
    String str = "";
    for (int j = 0; j < i; j++) {
      str+=c;        
    }
    return str;
  }
 
  // Driver Code
  public static void Main()
  {
    String s = "ilovegeeksforgeeks";
    int[] space = { 1, 5, 10, 13 };
 
    // Function Call
    Console.Write(spaceintegration(s, space) +"\n");
 
  }
}
 
// This code is contributed by Saurabh Jaiswal

Javascript

<script>
      // JavaScript code for the above approach
 
      // Function to add space in required positions
      function spaceintegration(s, sp)
      {
          let M = s.length, N = sp.length, l = 0, r = 0;
          let res = new Array(M + N).fill(' ');
 
          // Iterate over M+N length
          for (let i = 0; i < M + N; i++) {
 
              if (l < N && i == sp[l] + l)
                  l++;
              else
                  res[i] = s[r++];
          }
 
          // Return the required string
          return res.join('');
      }
 
      // Driver Code
      let s = "ilovegeeksforgeeks";
      let space = [1, 5, 10, 13];
 
      // Function Call
      document.write(spaceintegration(s, space) + '<br>');
 
// This code is contributed by Potta Lokesh
  </script>

Output

i love geeks for geeks

Time Complexity: O(M+N)
Auxiliary Space: O(M+N)

Approach#2: This problem can be solve by method which we use to insert element at specific position in array. Follow the steps below to solve the given problem.

We have string s and position in vector space.
Iterate over the element of the space From last of vector to first and follow the following steps for every element of vector. Let L element of the vector.
- Add one space at the end of s.
- Iterate over string Till the L:
  - Move one-one character forward till L.
- At last Add space at L.
Print string at the end.

Below is the implementation of the above approach.

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to add space in required positions
string spaceintegration(string s, vector<int>& space)
{
    int y = 0;
    int len = space.size();
 
    while (len--) {
        int k = space[len] + 1;
        int l = s.size() - 1;
        string tem = " ";
        s += tem;
        //   iterate over string
        for (int i = l; i >= k - 1; i--) {
            s[i + 1] = s[i];
        }
        s[k - 1] = tem[0];
        y += 1;
    }
    return s;
}
 
// Driver code
int main()
{
 
    string s = "ilovegeeksforgeeks";
    vector<int> space = { 1, 5, 10, 13 };
 
    //   Function call
    cout << spaceintegration(s, space) << endl;
 
    return 0;
}

Java

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
 
class GFG {
    // Function to add space in required positions
    public static String spaceIntegration(String s, List<Integer> space) {
        int len = space.size();
        while (len-- > 0) {
            int k = space.get(len) + 1;
            String tem = " ";
            s = s.substring(0, k - 1) + tem + s.substring(k - 1);
        }
        return s;
    }
 
    // Driver code
    public static void main(String[] args) {
        String s = "GeeksForGeeK";
        List<Integer> space = new ArrayList<Integer>();
        Collections.addAll(space, 1, 5, 10);
 
        // Function call
        System.out.println(spaceIntegration(s, space));
    }
}

Python

# Python3 program for above approach
 
# Function to add space in required positions
 
 
def spaceintegration(se, space):
    s = list(se)
    # Iterate over the string
    for i in range(len(space)-1, -1, -1):
        s.insert(space[i], " ")
 
    return "".join(s)
 
 
# Driver Code
if __name__ == "__main__":
 
    s = "ilovegeeksforgeeks"
 
    space = [1, 5, 10, 13]
 
    # Function Call
    print(spaceintegration(s, space))
 
# This code is contributed by ukasp

C#

// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to add space in required positions
    static string spaceintegration(string s, List<int> space)
    {
        int y = 0;
        int len = space.Count;
 
        while (len > 0) {
            len -= 1;
            int k = space[len] + 1;
            int l = s.Length - 1;
            string tem = " ";
            s += tem;
 
            // iterate over string
            for (int i = l - 1 ; i >= k - 1 ; i--) {
                  // Replaces s[i + 1] with s[i]
                s = s.Remove(i + 1, 1);
                s = s.Insert(i + 1, Char.ToString(s[i]));
            }
              // Replaces s[k - 1] with tem[0]
            s = s.Remove(k - 1, 1);
            s = s.Insert(k - 1, Char.ToString(tem[0]));
            y += 1;
        }
        return s;
    }
 
    // Driver code
    public static void Main(string[] args){
 
        string s = "ilovegeeksforgeeks";
        List<int> space = new List<int>{ 1, 5, 10, 13 };
 
        // Function call
        Console.WriteLine(spaceintegration(s, space));
 
    }
}
 
// This code is contributed by subhamgoyal2014.

Javascript

// JavaScript code for the above approach
 
        // Function to add space in required positions
        function spaceintegration(s, sp)
        {
           s = s.split('')
        //   iterate over the space
           for(let i = sp.length-1; i>=0; i--){
               s.splice(sp[i], 0, " ");
           }
 
            // Return the required string
            return s.join('');
        }
 
        // Driver Code
        let s = "ilovegeeksforgeeks";
        let space = [1, 5, 10, 13];
 
        // Function Call
        console.log(spaceintegration(s,space));
 
  // This code is contributed by Sam snehil

Output

i love geeks for geeks

Time Complexity: O(M*N) Here M is the length of the string and N is the size of space vector.
Auxiliary Space: O(1), As constant extra space is used.