How to give space in Java

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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++

#include <bits/stdc++.h>

using namespace std;

string spaceintegration(string s, vector<int>& sp)

{

    int M = s.size(), N = sp.size(), l = 0, r = 0;

    string res(M + N, ' ');

    for (int i = 0; i < M + N; i++) {

        if (l < N and i == sp[l] + l)

            l++;

        else

            res[i] = s[r++];

    }

    return res;

}

int main()

{

    string s = "ilovegeeksforgeeks";

    vector<int> space = { 1, 5, 10, 13 };

    cout << spaceintegration(s, space) << endl;

    return 0;

}

Java

import java.util.*;

class GFG

{

  static String spaceintegration(String s, int []sp)

  {

    int M = s.length(), N = sp.length, l = 0, r = 0;

    String res = newstr(M + N, ' ');

    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 res;

  }

  static String newstr(int i, char c) {

    String str = "";

    for (int j = 0; j < i; j++) {

      str+=c;       

    }

    return str;

  }

  public static void main(String[] args)

  {

    String s = "ilovegeeksforgeeks";

    int[] space = { 1, 5, 10, 13 };

    System.out.print(spaceintegration(s, space) +"\n");

  }

}

Python3

def spaceintegration(s, sp):

    M = len(s)

    N = len(sp)

    l = 0

    r = 0

    res = [' '] * (M + N)

    for i in range(M + N):

        if (l < N and i == sp[l] + l):

            l += 1

        else:

            res[i] = s[r]

            r += 1

    return ''.join(res)

if __name__ == "__main__":

    s = "ilovegeeksforgeeks"

    space = [ 1, 5, 10, 13 ]

    print(spaceintegration(s, space))

C#

using System;

class GFG

{

  static String spaceintegration(String s, int []sp)

  {

    int M = s.Length, N = sp.Length, l = 0, r = 0;

    String res = newstr(M + N, ' ');

    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 res;

  }

  static String newstr(int i, char c) {

    String str = "";

    for (int j = 0; j < i; j++) {

      str+=c;       

    }

    return str;

  }

  public static void Main()

  {

    String s = "ilovegeeksforgeeks";

    int[] space = { 1, 5, 10, 13 };

    Console.Write(spaceintegration(s, space) +"\n");

  }

}

Javascript

<script>

      function spaceintegration(s, sp)

      {

          let M = s.length, N = sp.length, l = 0, r = 0;

          let res = new Array(M + N).fill(' ');

          for (let i = 0; i < M + N; i++) {

              if (l < N && i == sp[l] + l)

                  l++;

              else

                  res[i] = s[r++];

          }

          return res.join('');

      }

      let s = "ilovegeeksforgeeks";

      let space = [1, 5, 10, 13];

      document.write(spaceintegration(s, space) + '<br>');

  </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.

Bellow is the implementation of above approach.

C++

#include <bits/stdc++.h>

using namespace std;

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;

        for (int i = l; i >= k - 1; i--) {

            s[i + 1] = s[i];

        }

        s[k - 1] = tem[0];

        y += 1;

    }

    return s;

}

int main()

{

    string s = "ilovegeeksforgeeks";

    vector<int> space = { 1, 5, 10, 13 };

    cout << spaceintegration(s, space) << endl;

    return 0;

}

Python

def spaceintegration(se, space):

    s = list(se)

    for i in range(len(space)-1, -1, -1):

        s.insert(space[i], " ")

    return "".join(s)

if __name__ == "__main__":

    s = "ilovegeeksforgeeks"

    space = [1, 5, 10, 13]

    print(spaceintegration(s, space))

C#

using System;

using System.Collections;

using System.Collections.Generic;

class GFG

{

    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;

            for (int i = l - 1 ; i >= k - 1 ; i--) {

                s = s.Remove(i + 1, 1);

                s = s.Insert(i + 1, Char.ToString(s[i]));

            }

            s = s.Remove(k - 1, 1);

            s = s.Insert(k - 1, Char.ToString(tem[0]));

            y += 1;

        }

        return s;

    }

    public static void Main(string[] args){

        string s = "ilovegeeksforgeeks";

        List<int> space = new List<int>{ 1, 5, 10, 13 };

        Console.WriteLine(spaceintegration(s, space));

    }

}

Javascript

        function spaceintegration(s, sp)

        {

           s = s.split('')

           for(let i = sp.length-1; i>=0; i--){

               s.splice(sp[i], 0, " ");

           }

            return s.join('');

        }

        let s = "ilovegeeksforgeeks";

        let space = [1, 5, 10, 13];

        console.log(spaceintegration(s,space));

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.