Subarray/Substring
A subarray is a contiguous part of 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 starting element and inner loop considers all elements on right of the picked elements as ending element of 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
// 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]+" "); } } } // Driver method to test the above function public static void main(String[] args) { System.out.println("All Non-empty Subarrays"); subArray(arr.length); } } |
Python3
# 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. |
C#
// 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. |
PHP
<?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 ?> |
Javascript
<script> // Javascript program to generate all // possible subarrays/subArrays // Complexity- O(n^3) let arr = [1, 2, 3, 4]; // Prints all subarrays in arr[0..n-1] function subArray(n) { // Pick starting point for(let i = 0; i < n; i++) { // Pick ending point for(let j = i; j < n; j++) { // Print subarray between current // starting and ending points for(let k = i; k <= j; k++) document.write(arr[k] + " "); document.write("</br>"); } } } // Driver code document.write("All Non-empty Subarrays" + "</br>"); subArray(arr.length); // This code is contributed by suresh07 </script> |
Output:
All 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)
Space Complexity: 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 “neveropen” and “gks”. “gks” is a subsequence of “neveropen” but not a substring. “neveropen” 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
/* 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); } } |
Python3
# 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. |
C#
// 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 |
PHP
<?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 ?> |
Javascript
<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> |
Output:
All 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))
Space Complexity: 0(1)
This article is contributed by Harshit Gupta. If you like neveropen and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
