Find minimum length sub-array which has given sub-sequence in it
Given an array arr[] of N elements, the task is to find the length of the smallest sub-array which has the sequence {0, 1, 2, 3, 4} as a sub-sequence in it.
Examples:
Input: arr[] = {0, 1, 2, 3, 4, 2, 0, 3, 4}
Output: 5
The required Subarray is {0, 1, 2, 3, 4} with minimum length.
The entire array also includes the sequence
but it is not minimum in length.Input: arr[] = {0, 1, 1, 0, 1, 2, 0, 3, 4}
Output: 6
Approach:
- Maintain an array pref[] of size 5 (equal to the size of the sequence) where pref[i] stores the count of i in the given array till now.
- We can increase the count of pref for any number only if pref[Array[i] – 1] > 0. This is because, in order to have the complete sequence as a sub-sequence of the array, all the previous elements of the sequence must occur before the current. Also, store the indices of these elements found so far.
- Whenever we witness 4 i.e., the possible end of the sub-sequence and pref[3] > 0 implies that we have found the sequence in our array. Now mark that index as the end as well as the start point and for all other numbers in sequence from 3 to 0. Apply binary search to find the closest index to the next element of the sequence, which will give us the size of the current valid sub-array.
- The answer is the minimum size of all the valid sub-arrays found in the previous step.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX_INT 1000000// Function to return the minimum length// of a sub-array which contains// {0, 1, 2, 3, 4} as a sub-sequenceint solve(int Array[], int N){ // To store the indices where 0, 1, 2, // 3 and 4 are present vector<int> pos[5]; // To store if there exist a valid prefix // of sequence in array int pref[5] = { 0 }; // Base Case if (Array[0] == 0) { pref[0] = 1; pos[0].push_back(0); } int ans = MAX_INT; for (int i = 1; i < N; i++) { // If current element is 0 if (Array[i] == 0) { // Update the count of 0s till now pref[0]++; // Push the index of the new 0 pos[0].push_back(i); } else { // To check if previous element of the // given sequence is found till now if (pref[Array[i] - 1] > 0) { pref[Array[i]]++; pos[Array[i]].push_back(i); // If it is the end of sequence if (Array[i] == 4) { int end = i; int start = i; // Iterate for other elements of the // sequence for (int j = 3; j >= 0; j--) { int s = 0; int e = pos[j].size() - 1; int temp = -1; // Binary Search to find closest // occurrence less than equal to // starting point while (s <= e) { int m = (s + e) / 2; if (pos[j][m] <= start) { temp = pos[j][m]; s = m + 1; } else { e = m - 1; } } // Update the starting point start = temp; } ans = min(ans, end - start + 1); } } } } return ans;}// Driver codeint main(){ int Array[] = { 0, 1, 2, 3, 4, 2, 0, 3, 4 }; int N = sizeof(Array) / sizeof(Array[0]); cout << solve(Array, N); return 0;} |
Java
// Java implementation of the approachclass GFG { static int MAX_INT = 1000000; // Function to return the minimum length // of a sub-array which contains // {0, 1, 2, 3, 4} as a sub-sequence static int solve(int[] array, int N) { // To store the indices where 0, 1, 2, // 3 and 4 are present int[][] pos = new int[5][10000]; // To store if there exist a valid prefix // of sequence in array int[] pref = new int[5]; // Base Case if (array[0] == 0) { pref[0] = 1; pos[0][pos[0].length - 1] = 0; } int ans = MAX_INT; for (int i = 1; i < N; i++) { // If current element is 0 if (array[i] == 0) { // Update the count of 0s till now pref[0]++; // Push the index of the new 0 pos[0][pos[0].length - 1] = i; } else { // To check if previous element of the // given sequence is found till now if (pref[array[i] - 1] > 0) { pref[array[i]]++; pos[array[i]][pos[array[i]].length - 1] = i; // If it is the end of sequence if (array[i] == 4) { int end = i; int start = i; // Iterate for other elements of the // sequence for (int j = 3; j >= 0; j--) { int s = 0; int e = pos[j].length - 1; int temp = -1; // Binary Search to find closest // occurrence less than equal to // starting point while (s <= e) { int m = (s + e) / 2; if (pos[j][m] <= start) { temp = pos[j][m]; s = m + 1; } else e = m - 1; } // Update the starting point start = temp; } ans = Math.min(ans, end - start + 1); } } } } return ans; } // Driver Code public static void main(String[] args) { int[] array = { 0, 1, 2, 3, 4, 2, 0, 3, 4 }; int N = array.length; System.out.println(solve(array, N)); }}// This code is contributed by// sanjeev2552 |
Python3
# Python3 implementation of the approachMAX_INT = 1000000# Function to return the minimum length# of a sub-array which contains# 0, 1, 2, 3, 4 as a sub-sequencedef solve(Array, N): # To store the indices where 0, 1, 2, # 3 and 4 are present pos = [[] for i in range(5)] # To store if there exist a valid prefix # of sequence in array pref = [0 for i in range(5)] # Base Case if (Array[0] == 0): pref[0] = 1 pos[0].append(0) ans = MAX_INT for i in range(N): # If current element is 0 if (Array[i] == 0): # Update the count of 0s till now pref[0] += 1 # Push the index of the new 0 pos[0].append(i) else: # To check if previous element of the # given sequence is found till now if (pref[Array[i] - 1] > 0): pref[Array[i]] += 1 pos[Array[i]].append(i) # If it is the end of sequence if (Array[i] == 4): end = i start = i # Iterate for other elements of the sequence for j in range(3, -1, -1): s = 0 e = len(pos[j]) - 1 temp = -1 # Binary Search to find closest occurrence # less than equal to starting point while (s <= e): m = (s + e) // 2 if (pos[j][m] <= start): temp = pos[j][m] s = m + 1 else: e = m - 1 # Update the starting point start = temp ans = min(ans, end - start + 1) return ans# Driver codeArray = [0, 1, 2, 3, 4, 2, 0, 3, 4]N = len(Array)print(solve(Array, N))# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approachusing System;class GFG { static int MAX_INT = 1000000; // Function to return the minimum length // of a sub-array which contains // {0, 1, 2, 3, 4} as a sub-sequence static int solve(int[] array, int N) { // To store the indices where 0, 1, 2, // 3 and 4 are present int[, ] pos = new int[5, 10000]; // To store if there exist a valid prefix // of sequence in array int[] pref = new int[5]; // Base Case if (array[0] == 0) { pref[0] = 1; pos[0, pos.GetLength(0) - 1] = 0; } int ans = MAX_INT; for (int i = 1; i < N; i++) { // If current element is 0 if (array[i] == 0) { // Update the count of 0s till now pref[0]++; // Push the index of the new 0 pos[0, pos.GetLength(0) - 1] = i; } else { // To check if previous element of the // given sequence is found till now if (pref[array[i] - 1] > 0) { pref[array[i]]++; pos[array[i], pos.GetLength(1) - 1] = i; // If it is the end of sequence if (array[i] == 4) { int end = i; int start = i; // Iterate for other elements of the // sequence for (int j = 3; j >= 0; j--) { int s = 0; int e = pos.GetLength(1) - 1; int temp = -1; // Binary Search to find closest // occurrence less than equal to // starting point while (s <= e) { int m = (s + e) / 2; if (pos[j, m] <= start) { temp = pos[j, m]; s = m + 1; } else e = m - 1; } // Update the starting point start = temp; } ans = Math.Min(ans, end - start + 1); } } } } return ans; } // Driver Code public static void Main(String[] args) { int[] array = { 0, 1, 2, 3, 4, 2, 0, 3, 4 }; int N = array.Length; Console.WriteLine(solve(array, N)); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the approachlet MAX_INT = 1000000;// Function to return the minimum length// of a sub-array which contains// {0, 1, 2, 3, 4} as a sub-sequencefunction solve(array,N){ // To store the indices where 0, 1, 2, // 3 and 4 are present let pos = new Array(5); for(let i=0;i<5;i++) { pos[i]=new Array(10000); for(let j=0;j<10000;j++) { pos[i][j]=0; } } // To store if there exist a valid prefix // of sequence in array let pref = new Array(5); for(let i=0;i<5;i++) { pref[i]=0; } // Base Case if (array[0] == 0) { pref[0] = 1; pos[0][pos[0].length - 1] = 0; } let ans = MAX_INT; for (let i = 1; i < N; i++) { // If current element is 0 if (array[i] == 0) { // Update the count of 0s till now pref[0]++; // Push the index of the new 0 pos[0][pos[0].length - 1] = i; } else { // To check if previous element of the // given sequence is found till now if (pref[array[i] - 1] > 0) { pref[array[i]]++; pos[array[i]][pos[array[i]].length - 1] = i; // If it is the end of sequence if (array[i] == 4) { let end = i; let start = i; // Iterate for other elements of the sequence for (let j = 3; j >= 0; j--) { let s = 0; let e = pos[j].length - 1; let temp = -1; // Binary Search to find closest occurrence // less than equal to starting point while (s <= e) { let m = Math.floor((s + e) / 2); if (pos[j][m] <= start) { temp = pos[j][m]; s = m + 1; } else e = m - 1; } // Update the starting point start = temp; } ans = Math.min(ans, end - start + 1); } } } } return ans;}// Driver Codelet array=[0, 1, 2, 3, 4, 2, 0, 3, 4];let N = array.length;document.write(solve(array, N));// This code is contributed by patel2127</script> |
Output:
5
Time Complexity: O(n.log n)
Auxiliary Space: O(n)
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