All substrings of a given String

Last Updated : 13 Feb, 2026

Given a string s, containing lowercase alphabetical characters. The task is to print all non-empty substrings of the given string.

Examples : 

Input : s = "abc"
Output : "a", "ab", "abc", "b", "bc", "c"

Input : s = "ab"
Output : "a", "ab", "b"

Input : s = "a"
Output : "a"

Table of Content

[Approach 1] Using Iteration - O(n^3) Time and O(n^3) Space

The idea is to use two nested loops. The outer loop selects the starting index from 0 to n-1, and for each starting index, the inner loop selects the ending index from that position up to n-1.

C++ 
#include <iostream>
using namespace std;

// Function to find all substrings
vector<string> findSubstrings(string &s) {

    // to store all substrings
    vector<string> res;

    for(int i = 0; i < s.length(); i++) {
        for(int j = i; j < s.length(); j++) {
            
            // substr function takes starting index
            // and length as parameters
            res.push_back(s.substr(i, j-i+1));
        }
    }

    return res;
}

int main() {
    string s = "abc";
    vector<string> res = findSubstrings(s);
    for(auto i:res) {
        cout<< i <<" ";
    }
    return 0;
}
C 
#include <stdio.h>
#include <string.h>

// Function to find all substrings
void findSubstrings(char *s) {
    
    // to store all substrings
    int len = strlen(s);
    for (int i = 0; i < len; i++) {
        for (int j = i; j < len; j++) {
            for (int k = i; k <= j; k++) {
                putchar(s[k]);
            }
            printf(" ");
        }
    }
}

int main() {
    char s[] = "abc";
    findSubstrings(s);
    return 0;
}
Java 
import java.util.ArrayList;
import java.util.List;

public class GFG {
    public static List<String> findSubstrings(String s) {
        
        // to store all substrings
        List<String> res = new ArrayList<>();
        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j < s.length(); j++) {
                
                // substr function takes starting index
                // and ending index + 1 as parameters
                res.add(s.substring(i, j + 1));
            }
        }
        return res;
    }

    public static void main(String[] args) {
        String s = "abc";
        List<String> res = findSubstrings(s);
        for (String i : res) {
            System.out.print(i + " ");
        }
    }
}
Python 
def find_substrings(s):
    
    # to store all substrings
    res = []
    for i in range(len(s)):
        for j in range(i, len(s)):
            res.append(s[i:j+1])
    return res


def main():
    s = 'abc'
    res = find_substrings(s)
    for i in res:
        print(i, end=' ')


if __name__ == "__main__":
    main()
C# 
using System;
using System.Collections.Generic;

class GFG {
    public static List<string> FindSubstrings(string s) {
        
        List<string> res = new List<string>();
        
        for (int i = 0; i < s.Length; i++) {
            for (int j = i; j < s.Length; j++) {
                
                // substr function takes starting index
                // and length as parameters
                res.Add(s.Substring(i, j - i + 1));
            }
        }
        return res;
    }

    static void Main() {
        string s = "abc";
        List<string> res = FindSubstrings(s);
        foreach (var i in res) {
            Console.Write(i + " ");
        }
    }
}
JavaScript 
function findSubstrings(s) {
    
    // to store all substrings
    let res = [];
    for (let i = 0; i < s.length; i++) {
        for (let j = i; j < s.length; j++) {
            
            // substr function takes starting index
            // and ending index + 1 as parameters
            res.push(s.substring(i, j + 1));
        }
    }
    return res;
}

//Driver Code
let s = 'abc';
let res = findSubstrings(s);
res.forEach(i => {
    console.log(i + ' ');
});

Output
a ab abc b bc c

[Approach 2] Using Recursion - O(n^3) Time and O(n^3) Space

The idea is to use recursion to generate all substrings. Start from index 0 and build a current string cur by adding one character at a time. After adding each character, store cur in the result list. Then recursively move to the next index to continue building longer substrings. If cur becomes empty, skip the current character and start forming substrings from the next index.

C++ 
#include <iostream>
using namespace std;

// Recursive Function to find all
// substrings of a string
void subString(string &s, int n, int index, 
            string &cur, vector<string> &res) {

    // if we have reached the 
    // end of the string
    if (index == n) {
        return;
    }

    // add the character s[index] 
    // to the current string
    cur.push_back(s[index]);

    // add the current string in result
    res.push_back(cur);

    // move to next index
    subString(s, n, index + 1, cur, res);

    // remove the current character
    // from the current string 
    cur.pop_back();

    
    // if current string is empty
    // skip the current index to
    // start the new substring
    if(cur.empty()) {
        subString(s, n, index + 1, cur, res);
    }
}

// Function to find all substrings
vector<string> findSubstrings(string s) {

    // to store all substrings
    vector<string> res;
    
    // to store current string
    string cur = "";
    subString(s, s.length(), 0, cur, res);
    return res;
}

int main() {
    string s = "abc";
    vector<string> res = findSubstrings(s);
    for(auto i:res) {
        cout<< i <<" ";
    }
    return 0;
}
Java 
import java.util.*;

class GFG {

    // Recursive Function to find all
    // substrings of a string
    static void subString(String s, int n, int index,
                          StringBuilder cur, List<String> res) {

        // if we have reached the
        // end of the string
        if (index == n) {
            return;
        }

        // add the character s.charAt(index)
        // to the current string
        cur.append(s.charAt(index));

        // add the current string in result
        res.add(cur.toString());

        // move to next index
        subString(s, n, index + 1, cur, res);

        // remove the current character
        // from the current string
        cur.deleteCharAt(cur.length() - 1);

        // if current string is empty
        // skip the current index to
        // start the new substring
        if (cur.length() == 0) {
            subString(s, n, index + 1, cur, res);
        }
    }

    // Function to find all substrings
    static List<String> findSubstrings(String s) {

        // to store all substrings
        List<String> res = new ArrayList<>();

        // to store current string
        StringBuilder cur = new StringBuilder();
        subString(s, s.length(), 0, cur, res);
        return res;
    }

    public static void main(String[] args) {
        String s = "abc";
        List<String> res = findSubstrings(s);
        for (String str : res) {
            System.out.print(str + " ");
        }
    }
}
Python 
def subString(s, n, index, cur, res):

    # if we have reached the
    # end of the string
    if index == n:
        return

    # add the character s[index]
    # to the current string
    cur += s[index]

    # add the current string in result
    res.append(cur)

    # move to next index
    subString(s, n, index + 1, cur, res)

    # remove the current character
    # from the current string
    cur = cur[:-1]

    # if current string is empty
    # skip the current index to
    # start the new substring
    if not cur:
        subString(s, n, index + 1, cur, res)

# Function to find all substrings
def findSubstrings(s):

    # to store all substrings
    res = []

    # to store current string
    cur = ""
    subString(s, len(s), 0, cur, res)
    return res

if __name__ == "__main__":
    s = "abc"
    res = findSubstrings(s)
    print(" ".join(res))
C# 
using System;
using System.Collections.Generic;

class GFG {

    // Recursive Function to find all
    // substrings of a string
    static void subString(string s, int n, int index,
                          List<char> cur, List<string> res) {

        // if we have reached the
        // end of the string
        if (index == n) {
            return;
        }

        // add the character s[index]
        // to the current string
        cur.Add(s[index]);

        // add the current string in result
        res.Add(new string(cur.ToArray()));

        // move to next index
        subString(s, n, index + 1, cur, res);

        // remove the current character
        // from the current string
        cur.RemoveAt(cur.Count - 1);

        // if current string is empty
        // skip the current index to
        // start the new substring
        if (cur.Count == 0) {
            subString(s, n, index + 1, cur, res);
        }
    }

    // Function to find all substrings
    static List<string> findSubstrings(string s) {

        // to store all substrings
        List<string> res = new List<string>();

        // to store current string
        List<char> cur = new List<char>();
        subString(s, s.Length, 0, cur, res);
        return res;
    }

    public static void Main() {
        string s = "abc";
        List<string> res = findSubstrings(s);
        Console.WriteLine(string.Join(" ", res));
    }
}
JavaScript 
function subString(s, n, index, cur, res) {

    // if we have reached the
    // end of the string
    if (index === n) {
        return;
    }

    // add the character s[index]
    // to the current string
    cur += s[index];

    // add the current string in result
    res.push(cur);

    // move to next index
    subString(s, n, index + 1, cur, res);

    // remove the current character
    // from the current string
    cur = cur.slice(0, -1);

    // if current string is empty
    // skip the current index to
    // start the new substring
    if (cur.length === 0) {
        subString(s, n, index + 1, cur, res);
    }
}

// Function to find all substrings
function findSubstrings(s) {

    // to store all substrings
    let res = [];

    // to store current string
    let cur = "";
    subString(s, s.length, 0, cur, res);
    return res;
}

//Driver Code
const s = "abc";
const res = findSubstrings(s);
console.log(res.join(" "));

Output
a ab abc b bc c
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Article Tags:
Strings
DSA
Basic Coding Problems
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