Subarray/Substring vs Subsequence and Programs to Generate them
Last Updated :
23 Jul, 2025
Subarray/Substring
A subarray is a contiguous part of the array. An array that is inside another array. For example, consider the array [1, 2, 3, 4], There are 10 non-empty sub-arrays. The subarrays are (1), (2), (3), (4), (1,2), (2,3), (3,4), (1,2,3), (2,3,4) and (1,2,3,4). In general, for an array/string of size n, there are n*(n+1)/2 non-empty subarrays/substrings.
How to generate all subarrays?
We can run two nested loops, the outer loop picks the starting element and the inner loop considers all elements on the right of the picked elements as the ending elements of the subarray.
C++
/* C++ code to generate all possible subarrays/subArrays
Complexity- O(n^3) */
#include<bits/stdc++.h>
using namespace std;
// Prints all subarrays in arr[0..n-1]
void subArray(int arr[], int n)
{
// Pick starting point
for (int i=0; i <n; i++)
{
// Pick ending point
for (int j=i; j<n; j++)
{
// Print subarray between current starting
// and ending points
for (int k=i; k<=j; k++)
cout << arr[k] << " ";
cout << endl;
}
}
}
// Driver program
int main()
{
int arr[] = {1, 2, 3, 4};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "All Non-empty Subarrays\n";
subArray(arr, n);
return 0;
}
// Java program to generate all possible subarrays/subArrays
// Complexity- O(n^3) */
class Test
{
static int arr[] = new int[]{1, 2, 3, 4};
// Prints all subarrays in arr[0..n-1]
static void subArray( int n)
{
// Pick starting point
for (int i=0; i <n; i++)
{
// Pick ending point
for (int j=i; j<n; j++)
{
// Print subarray between current starting
// and ending points
for (int k=i; k<=j; k++)
System.out.print(arr[k]+" ");
System.out.println();
}
}
}
// Driver method to test the above function
public static void main(String[] args)
{
System.out.println("All Non-empty Subarrays");
subArray(arr.length);
}
}
// C# program to generate all
// possible subarrays/subArrays
// Complexity- O(n^3)
using System;
class GFG
{
static int []arr = new int[]{1, 2, 3, 4};
// Prints all subarrays in arr[0..n-1]
static void subArray( int n)
{
// Pick starting point
for (int i = 0; i < n; i++)
{
// Pick ending point
for (int j = i; j < n; j++)
{
// Print subarray between current
// starting and ending points
for (int k = i; k <= j; k++)
Console.Write(arr[k]+" ");
Console.WriteLine("");
}
}
}
// Driver Code
public static void Main()
{
Console.WriteLine("All Non-empty Subarrays");
subArray(arr.Length);
}
}
// This code is contributed by Sam007.
//Function that returns all subarrays
let getSubArr = (arr) => {
//Empty array to store subarrays
let arr1 = [];
for(let i=0;i<arr.length;i++) {
//Empty array to store one subarray
let subArr = [];
for(let j=i;j<arr.length;j++) {
//To get elements to the right of selected i th element till j
subArr.push(arr.slice(i,j+1));
}
//Storing individual subarrays into main array
arr1.push(subArr);
}
//Return the array of subarrays
return arr1;
}
//Example Array
let arr = [1,2,3,4];
let arr1 = getSubArr(arr);
console.log(arr1);
<?php
// PHP code to generate all possible
// subarrays/subArrays Complexity- O(n^3)
// Prints all subarrays
// in arr[0..n-1]
function subArray($arr, $n)
{
// Pick starting point
for ($i = 0; $i < $n; $i++)
{
// Pick ending point
for ($j = $i; $j < $n; $j++)
{
// Print subarray between
// current starting
// and ending points
for ($k = $i; $k <= $j; $k++)
echo $arr[$k] , " ";
echo "\n";
}
}
}
// Driver Code
$arr= array(1, 2, 3, 4);
$n = sizeof($arr);
echo "All Non-empty Subarrays\n";
subArray($arr, $n);
// This code is contributed by m_kit
?>
# Python3 code to generate all possible
# subarrays/subArrays
# Complexity- O(n^3)
# Prints all subarrays in arr[0..n-1]
def subArray(arr, n):
# Pick starting point
for i in range(0,n):
# Pick ending point
for j in range(i,n):
# Print subarray between
# current starting
# and ending points
for k in range(i,j+1):
print (arr[k],end=" ")
print ("\n",end="")
# Driver program
arr = [1, 2, 3, 4]
n = len(arr)
print ("All Non-empty Subarrays")
subArray(arr, n);
# This code is contributed by Shreyanshi.
OutputAll Non-empty Subarrays
1
1 2
1 2 3
1 2 3 4
2
2 3
2 3 4
3
3 4
4
Time Complexity: 0(n^3)
Auxiliary Space: 0(1)
Subsequence: A subsequence is a sequence that can be derived from another sequence by removing zero or more elements, without changing the order of the remaining elements.
For the same example, there are 15 sub-sequences. They are (1), (2), (3), (4), (1,2), (1,3),(1,4), (2,3), (2,4), (3,4), (1,2,3), (1,2,4), (1,3,4), (2,3,4), (1,2,3,4). More generally, we can say that for a sequence of size n, we can have (2n-1) non-empty sub-sequences in total.
A string example to differentiate: Consider strings "geeksforgeeks" and "gks". "gks" is a subsequence of "geeksforgeeks" but not a substring. "geeks" is both a subsequence and subarray. Every subarray is a subsequence. More specifically, Subsequence is a generalization of substring.
A subarray or substring will always be contiguous, but a subsequence need not be contiguous. That is, subsequences are not required to occupy consecutive positions within the original sequences. But we can say that both contiguous subsequence and subarray are the same.
How to generate all Subsequences?
We can use algorithm to generate power set for generation of all subsequences.
C++
/* C++ code to generate all possible subsequences.
Time Complexity O(n * 2^n) */
#include<bits/stdc++.h>
using namespace std;
void printSubsequences(int arr[], int n)
{
/* Number of subsequences is (2**n -1)*/
unsigned int opsize = pow(2, n);
/* Run from counter 000..1 to 111..1*/
for (int counter = 1; counter < opsize; counter++)
{
for (int j = 0; j < n; j++)
{
/* Check if jth bit in the counter is set
If set then print jth element from arr[] */
if (counter & (1<<j))
cout << arr[j] << " ";
}
cout << endl;
}
}
// Driver program
int main()
{
int arr[] = {1, 2, 3, 4};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "All Non-empty Subsequences\n";
printSubsequences(arr, n);
return 0;
}
/* Java code to generate all possible subsequences.
Time Complexity O(n * 2^n) */
import java.math.BigInteger;
class Test
{
static int arr[] = new int[]{1, 2, 3, 4};
static void printSubsequences(int n)
{
/* Number of subsequences is (2**n -1)*/
int opsize = (int)Math.pow(2, n);
/* Run from counter 000..1 to 111..1*/
for (int counter = 1; counter < opsize; counter++)
{
for (int j = 0; j < n; j++)
{
/* Check if jth bit in the counter is set
If set then print jth element from arr[] */
if (BigInteger.valueOf(counter).testBit(j))
System.out.print(arr[j]+" ");
}
System.out.println();
}
}
// Driver method to test the above function
public static void main(String[] args)
{
System.out.println("All Non-empty Subsequences");
printSubsequences(arr.length);
}
}
// C# code to generate all possible subsequences.
// Time Complexity O(n * 2^n)
using System;
class GFG{
static void printSubsequences(int[] arr, int n)
{
// Number of subsequences is (2**n -1)
int opsize = (int)Math.Pow(2, n);
// Run from counter 000..1 to 111..1
for(int counter = 1; counter < opsize; counter++)
{
for(int j = 0; j < n; j++)
{
// Check if jth bit in the counter is set
// If set then print jth element from arr[]
if ((counter & (1 << j)) != 0)
Console.Write(arr[j] + " ");
}
Console.WriteLine();
}
}
// Driver Code
static void Main()
{
int[] arr = { 1, 2, 3, 4 };
int n = arr.Length;
Console.WriteLine("All Non-empty Subsequences");
printSubsequences(arr, n);
}
}
// This code is contributed by divyesh072019
<script>
// Javascript code to generate all possible subsequences.
// Time Complexity O(n * 2^n)
function printSubsequences(arr, n)
{
// Number of subsequences is (2**n -1)
let opsize = parseInt(Math.pow(2, n), 10);
// Run from counter 000..1 to 111..1
for(let counter = 1; counter < opsize; counter++)
{
for(let j = 0; j < n; j++)
{
// Check if jth bit in the counter is set
// If set then print jth element from arr[]
if ((counter & (1 << j)) != 0)
document.write(arr[j] + " ");
}
document.write("</br>");
}
}
let arr = [ 1, 2, 3, 4 ];
let n = arr.length;
document.write("All Non-empty Subsequences" + "</br>");
printSubsequences(arr, n);
// This code is contributed by divyeshrabadiya07.
</script>
<?php
// PHP code to generate all
// possible subsequences.
// Time Complexity O(n * 2^n)
function printSubsequences($arr, $n)
{
// Number of subsequences
// is (2**n -1)
$opsize = pow(2, $n);
/* Run from counter
000..1 to 111..1*/
for ($counter = 1;
$counter < $opsize;
$counter++)
{
for ( $j = 0; $j < $n; $j++)
{
/* Check if jth bit in
the counter is set
If set then print jth
element from arr[] */
if ($counter & (1 << $j))
echo $arr[$j], " ";
}
echo "\n";
}
}
// Driver Code
$arr = array (1, 2, 3, 4);
$n = sizeof($arr);
echo "All Non-empty Subsequences\n";
printSubsequences($arr, $n);
// This code is contributed by ajit
?>
# Python3 code to generate all
# possible subsequences.
# Time Complexity O(n * 2 ^ n)
import math
def printSubsequences(arr, n) :
# Number of subsequences is (2**n -1)
opsize = math.pow(2, n)
# Run from counter 000..1 to 111..1
for counter in range( 1, (int)(opsize)) :
for j in range(0, n) :
# Check if jth bit in the counter
# is set If set then print jth
# element from arr[]
if (counter & (1<<j)) :
print( arr[j], end =" ")
print()
# Driver program
arr = [1, 2, 3, 4]
n = len(arr)
print( "All Non-empty Subsequences")
printSubsequences(arr, n)
# This code is contributed by Nikita Tiwari.
OutputAll Non-empty Subsequences
1
2
1 2
3
1 3
2 3
1 2 3
4
1 4
2 4
1 2 4
3 4
1 3 4
2 3 4
1 2 3 4
Time Complexity: 0(n*(2^n))
Auxiliary Space: 0(1)
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