Find All Occurrences of Subarray in Array
Last Updated : 26 Nov, 2024
Given two arrays a[] and b[], the task is to find all starting indices of b[] as a subarray in a[].
Examples:
Input: a[] = [2, 3, 0, 3, 0, 3, 0], b[] = [3, 0, 3, 0]
Output: [1, 3]
Explanation: The subarray a[1...4] = b[] and subarray a[3...6] = b[].
Input : a[] = [1, 2, 3, 4, 5], b[] = [2, 5, 6]
Output: []
Explanation: No subarray of a[] matches with b[].
[Naive Approach] Comparing All Subarrays - O(n*m) Time and O(1) Space
The idea is to check for all possible indices in a[] as starting index of subarray b[]. For each index, compare the subarray of a[] with b[] using a nested loop. If all the elements match, store the starting index in result. If any element does not match, break and check for next starting index.
C++ // C++ Program to search for subarray by matching // with every possible subarray #include <iostream> #include <vector> using namespace std; vector<int> search(vector<int> &a, vector<int> &b) { int n = a.size(), m = b.size(); vector<int> res; // Iterate over all possible starting indices for(int i = 0; i < n - m + 1; i++) { bool isSame = true; for(int j = 0; j < m; j++) { // If any character does not match, break // and begin from the next starting index if(a[i + j] != b[j]) { isSame = false; break; } } // If all characters are matched, store the // starting index if(isSame) res.push_back(i); } return res; } int main() { vector<int> a = {2, 3, 0, 3, 0, 3, 0}; vector<int> b = {3, 0, 3, 0}; vector<int> res = search(a, b); for(int idx: res) cout << idx << " "; } // Java Program to search for subarray by matching // with every possible subarray import java.util.ArrayList; import java.util.List; class GfG { static List<Integer> search(int[] a, int[] b) { int n = a.length, m = b.length; List<Integer> res = new ArrayList<>(); // Iterate over all possible starting indices for (int i = 0; i < n - m + 1; i++) { boolean isSame = true; // If any character does not match, break // and begin from the next starting index for (int j = 0; j < m; j++) { if (a[i + j] != b[j]) { isSame = false; break; } } // If all characters are matched, store // the starting index if (isSame) res.add(i); } return res; } public static void main(String[] args) { int[] a = {2, 3, 0, 3, 0, 3, 0}; int[] b = {3, 0, 3, 0}; List<Integer> res = search(a, b); for (int idx : res) System.out.print(idx + " "); } } # Python Program to search for subarray by matching # with every possible subarray def search(a, b): n = len(a) m = len(b) res = [] # Iterate over all possible starting indices for i in range(n - m + 1): isSame = True for j in range(m): # If any character does not match, break # and begin from the next starting index if a[i + j] != b[j]: isSame = False break # If all characters are matched, store the starting index if isSame: res.append(i) return res if __name__ == "__main__": a = [2, 3, 0, 3, 0, 3, 0] b = [3, 0, 3, 0] res = search(a, b) for idx in res: print(idx, end=" ")
// C# Program to search for subarray by matching // with every possible subarray using System; using System.Collections.Generic; class GfG { static List<int> Search(int[] a, int[] b) { int n = a.Length, m = b.Length; List<int> res = new List<int>(); // Iterate over all possible starting indices for (int i = 0; i < n - m + 1; i++) { bool isSame = true; for (int j = 0; j < m; j++) { // If any character does not match, break // and begin from the next starting index if (a[i + j] != b[j]) { isSame = false; break; } } // If all characters are matched, store the starting index if (isSame) res.Add(i); } return res; } static void Main() { int[] a = { 2, 3, 0, 3, 0, 3, 0 }; int[] b = { 3, 0, 3, 0 }; List<int> res = Search(a, b); foreach (int idx in res) { Console.Write(idx + " "); } } } // JavaScript Program to search for subarray by matching // with every possible subarray function search(a, b) { let n = a.length, m = b.length; let res = []; // Iterate over all possible starting indices for (let i = 0; i < n - m + 1; i++) { let isSame = true; for (let j = 0; j < m; j++) { // If any character does not match, break // and begin from the next starting index if (a[i + j] !== b[j]) { isSame = false; break; } } // If all characters are matched, store the starting index if (isSame) res.push(i); } return res; } // Driver code let a = [2, 3, 0, 3, 0, 3, 0]; let b = [3, 0, 3, 0]; let res = search(a, b); for (let idx of res) { console.log(idx + " "); } Time Complexity: O(n*m), where n and m are the sizes of the arrays a[] and b[], respectively.
Space Complexity: O(1) as we are not using any additional space to store the arrays or any other variables.
[Expected Approach] Using KMP Algorithm - O(n+m) Time and O(m) Space
The idea is to use KMP Algorithm with a[] as the text and b[] as the pattern. So, instead of comparing characters, we can compare numbers of the array to construct the lps[] array and find all occurrences of b[] in a[].
C++ // C++ Program to search for subarray using KMP Algorithm #include <iostream> #include <vector> using namespace std; void constructLps(vector<int> &pat, vector<int> &lps) { // len stores the length of longest prefix which // is also a suffix for the previous index int len = 0; // lps[0] is always 0 lps[0] = 0; int i = 1; while (i < pat.size()) { // If numbers match, increment the size of lps if (pat[i] == pat[len]) { len++; lps[i] = len; i++; } // If there is a mismatch else { if (len != 0) { // Update len to the previous lps value // to avoid reduntant comparisons len = lps[len - 1]; } else { // If no matching prefix found, set lps[i] to 0 lps[i] = 0; i++; } } } } vector<int> search(vector<int> &a, vector<int> &b) { int n = a.size(); int m = b.size(); vector<int> lps(m); vector<int> res; constructLps(b, lps); // Pointers i and j, for traversing a[] and b[] int i = 0; int j = 0; while (i < n) { // If elements match, move both pointers forward if (a[i] == b[j]) { i++; j++; // If all elements of b[] are matched // store the start index in result if (j == m) { res.push_back(i - j); // Use LPS of previous index to // skip unnecessary comparisons j = lps[j - 1]; } } // If there is a mismatch else { // Use lps value of previous index // to avoid redundant comparisons if (j != 0) j = lps[j - 1]; else i++; } } return res; } int main() { vector<int> a = {2, 3, 0, 3, 0, 3, 0}; vector<int> b = {3, 0, 3, 0}; vector<int> res = search(a, b); for(int idx: res) cout << idx << " "; } // Java Program to search for subarray using KMP Algorithm import java.util.ArrayList; import java.util.List; class GfG { // Function to construct LPS array static void constructLps(int[] pat, int[] lps) { // len stores the length of longest prefix which // is also a suffix for the previous index int len = 0; // lps[0] is always 0 lps[0] = 0; int i = 1; while (i < pat.length) { // If numbers match, increment the size of lps if (pat[i] == pat[len]) { len++; lps[i] = len; i++; } // If there is a mismatch else { if (len != 0) { // Update len to the previous lps value // to avoid redundant comparisons len = lps[len - 1]; } else { // If no matching prefix found, set lps[i] to 0 lps[i] = 0; i++; } } } } // Function to search for the subarray using KMP algorithm static List<Integer> search(int[] a, int[] b) { int n = a.length; int m = b.length; int[] lps = new int[m]; List<Integer> res = new ArrayList<>(); constructLps(b, lps); // Pointers i and j, for traversing a[] and b[] int i = 0; int j = 0; while (i < n) { // If elements match, move both pointers forward if (a[i] == b[j]) { i++; j++; // If all elements of b[] are matched // store the start index in result if (j == m) { res.add(i - j); // Use LPS of previous index to // skip unnecessary comparisons j = lps[j - 1]; } } // If there is a mismatch else { // Use lps value of previous index // to avoid redundant comparisons if (j != 0) j = lps[j - 1]; else i++; } } return res; } public static void main(String[] args) { int[] a = {2, 3, 0, 3, 0, 3, 0}; int[] b = {3, 0, 3, 0}; List<Integer> res = search(a, b); for (int idx : res) System.out.print(idx + " "); } } # Python Program to search for subarray using KMP Algorithm def constructLps(pat, lps): # len stores the length of longest prefix which # is also a suffix for the previous index length = 0 # lps[0] is always 0 lps[0] = 0 i = 1 while i < len(pat): # If numbers match, increment the size of lps if pat[i] == pat[length]: length += 1 lps[i] = length i += 1 # If there is a mismatch else: if length != 0: # Update length to the previous lps value # to avoid redundant comparisons length = lps[length - 1] else: # If no matching prefix found, set lps[i] to 0 lps[i] = 0 i += 1 def search(a, b): n = len(a) m = len(b) lps = [0] * m res = [] constructLps(b, lps) # Pointers i and j, for traversing a[] and b[] i = 0 j = 0 while i < n: # If elements match, move both pointers forward if a[i] == b[j]: i += 1 j += 1 # If all elements of b[] are matched # store the start index in result if j == m: res.append(i - j) # Use LPS of previous index to # skip unnecessary comparisons j = lps[j - 1] else: # If there is a mismatch # Use lps value of previous index # to avoid redundant comparisons if j != 0: j = lps[j - 1] else: i += 1 return res if __name__ == "__main__": a = [2, 3, 0, 3, 0, 3, 0] b = [3, 0, 3, 0] res = search(a, b) for idx in res: print(idx, end=" ")
// C# Program to search for subarray using KMP Algorithm using System; using System.Collections.Generic; class GfG { // Function to construct the LPS array (Longest Prefix Suffix) static void ConstructLps(int[] pat, int[] lps) { // len stores the length of the longest prefix which // is also a suffix for the previous index int len = 0; // lps[0] is always 0 lps[0] = 0; int i = 1; while (i < pat.Length) { // If numbers match, increment the size of lps if (pat[i] == pat[len]) { len++; lps[i] = len; i++; } // If there is a mismatch else { if (len != 0) { // Update len to the previous lps value // to avoid redundant comparisons len = lps[len - 1]; } else { // If no matching prefix found, set lps[i] to 0 lps[i] = 0; i++; } } } } // Function to search for the subarray static List<int> Search(int[] a, int[] b) { int n = a.Length; int m = b.Length; int[] lps = new int[m]; List<int> res = new List<int>(); ConstructLps(b, lps); // Pointers i and j, for traversing a[] and b[] int i = 0; int j = 0; while (i < n) { // If elements match, move both pointers forward if (a[i] == b[j]) { i++; j++; // If all elements of b[] are matched // store the start index in result if (j == m) { res.Add(i - j); // Use LPS of previous index to // skip unnecessary comparisons j = lps[j - 1]; } } // If there is a mismatch else { // Use lps value of previous index // to avoid redundant comparisons if (j != 0) j = lps[j - 1]; else i++; } } // Convert the List<int> to an int[] before returning return res; } static void Main() { int[] a = { 2, 3, 0, 3, 0, 3, 0 }; int[] b = { 3, 0, 3, 0 }; List<int> res = Search(a, b); foreach (int idx in res) { Console.Write(idx + " "); } } } // JavaScript Program to search for subarray using KMP Algorithm function constructLps(pat, lps) { // len stores the length of longest prefix which // is also a suffix for the previous index let len = 0; // lps[0] is always 0 lps[0] = 0; let i = 1; while (i < pat.length) { // If numbers match, increment the size of lps if (pat[i] === pat[len]) { len++; lps[i] = len; i++; } // If there is a mismatch else { if (len !== 0) { // Update len to the previous lps value // to avoid redundant comparisons len = lps[len - 1]; } else { // If no matching prefix found, set lps[i] to 0 lps[i] = 0; i++; } } } } function search(a, b) { let n = a.length; let m = b.length; let lps = new Array(m); let res = []; constructLps(b, lps); // Pointers i and j, for traversing a[] and b[] let i = 0; let j = 0; while (i < n) { // If elements match, move both pointers forward if (a[i] === b[j]) { i++; j++; // If all elements of b[] are matched // store the start index in result if (j === m) { res.push(i - j); // Use LPS of previous index to // skip unnecessary comparisons j = lps[j - 1]; } } // If there is a mismatch else { // Use lps value of previous index // to avoid redundant comparisons if (j !== 0) j = lps[j - 1]; else i++; } } return res; } // Driver Code let a = [2, 3, 0, 3, 0, 3, 0]; let b = [3, 0, 3, 0]; let res = search(a, b); for (let idx of res) { console.log(idx + " "); } Time Complexity: O(n+m), where n and m are the sizes of the arrays a[] and b[], respectively.
Auxiliary Space: O(m), for lps[] array.
Related Topic: Subarrays, Subsequences, and Subsets in Array
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