Given an array arr of N elements, the task is to find the length of the smallest subarray of the given array that contains at least one duplicate element. A subarray is formed from consecutive elements of an array. If no such array exists, print “-1”.
Examples:
Input: arr = {1, 2, 3, 1, 5, 4, 5}
Output: 3
Explanation:
Input: arr = {4, 7, 11, 3, 1, 2, 4}
Output: 7
Explanation:
Naive Approach:
- The trick is to find all pairs of two elements with equal value. Since these two elements have equal value, the subarray enclosing them would have at least a single duplicate and will be one of the candidates for the answer.
- A simple solution is to use two nested loops to find every pair of elements.If the two elements are equal then update the maximum length obtained so far.
Below is the implementation of the above approach:
C++
// C++ program to find// the smallest subarray having// atleast one duplicate#include <bits/stdc++.h>using namespace std;// Function to calculate// SubArray Lengthint subArrayLength(int arr[], int n){ int minLen = INT_MAX; for (int i = 1; i < n; i++) { for (int j = 0; j < i; j++) { // If the two elements are equal, // then the subarray arr[i..j] // will definitely have a duplicate if (arr[i] == arr[j]) { // Update the minimum length // obtained so far minLen = min(minLen, i - j + 1); } } } if (minLen == INT_MAX) { return -1; } return minLen;}// Driver Codeint main(){ int n = 7; int arr[] = { 1, 2, 3, 1, 5, 4, 5 }; int ans = subArrayLength(arr, n); cout << ans << '\n'; return 0;} |
Java
// Java program to find // the smallest subarray having // atleast one duplicateclass GFG{ final static int INT_MAX = Integer.MAX_VALUE; // Function to calculate // SubArray Length static int subArrayLength(int arr[], int n) { int minLen = INT_MAX; for (int i = 1; i < n; i++) { for (int j = 0; j < i; j++) { // If the two elements are equal, // then the subarray arr[i..j] // will definitely have a duplicate if (arr[i] == arr[j]) { // Update the minimum length // obtained so far minLen = Math.min(minLen, i - j + 1); } } } if (minLen == INT_MAX) { return -1; } return minLen; } // Driver Code public static void main(String[] args) { int n = 7; int arr[] = { 1, 2, 3, 1, 5, 4, 5 }; int ans = subArrayLength(arr, n); System.out.println(ans); } }// This code is contributed by AnkitRai01 |
Python
# Python program for above approachn = 7arr = [1, 2, 3, 1, 5, 4, 5]minLen = n + 1for i in range(1, n): for j in range(0, i): if arr[i]== arr[j]: minLen = min(minLen, i-j + 1)if minLen == n + 1: print("-1")else: print(minLen) |
C#
// C# program to find // the smallest subarray having // atleast one duplicateusing System;class GFG{ static int INT_MAX = int.MaxValue; // Function to calculate // SubArray Length static int subArrayLength(int []arr, int n) { int minLen = INT_MAX; for (int i = 1; i < n; i++) { for (int j = 0; j < i; j++) { // If the two elements are equal, // then the subarray arr[i..j] // will definitely have a duplicate if (arr[i] == arr[j]) { // Update the minimum length // obtained so far minLen = Math.Min(minLen, i - j + 1); } } } if (minLen == INT_MAX) { return -1; } return minLen; } // Driver Code public static void Main() { int n = 7; int []arr = { 1, 2, 3, 1, 5, 4, 5 }; int ans = subArrayLength(arr, n); Console.WriteLine(ans); } }// This code is contributed by AnkitRai01 |
Javascript
<script>// javascript program to find // the smallest subarray having // atleast one duplicate var INT_MAX = Number.MAX_VALUE; // Function to calculate // SubArray Length function subArrayLength( arr , n) { var minLen = INT_MAX; for (var i = 1; i < n; i++) { for (var j = 0; j < i; j++) { // If the two elements are equal, // then the subarray arr[i..j] // will definitely have a duplicate if (arr[i] == arr[j]) { // Update the minimum length // obtained so far minLen = Math.min(minLen, i - j + 1); } } } if (minLen == INT_MAX) { return -1; } return minLen; } // Driver Code var n = 7; var arr = [ 1, 2, 3, 1, 5, 4, 5 ]; var ans = subArrayLength(arr, n); document.write(ans);// This code contributed by Princi Singh</script> |
Output:
3
Time Complexity: O(N2)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient Approach:
This problem can be solved in O(N) time and O(N) Auxiliary space using the idea of hashing technique. The idea is to iterate through each element of the array in a linear way and for each element, find its last occurrence using a hashmap and then update the value of min length using the difference of the last occurrence and the current index. Also, update the value of the last occurrence of the element by the value of the current index.
Below is the implementation of the above approach:
C++
// C++ program to find// the smallest subarray having// atleast one duplicate#include <bits/stdc++.h>using namespace std;// Function to calculate// SubArray Lengthint subArrayLength(int arr[], int n){ int minLen = INT_MAX; // Last stores the index of the last // occurrence of the corresponding value unordered_map<int, int> last; for (int i = 0; i < n; i++) { // If the element has already occurred if (last[arr[i]] != 0) { minLen = min(minLen, i - last[arr[i]] + 2); } last[arr[i]] = i + 1; } if (minLen == INT_MAX) { return -1; } return minLen;}// Driver Codeint main(){ int n = 7; int arr[] = { 1, 2, 3, 1, 5, 4, 5 }; int ans = subArrayLength(arr, n); cout << ans << '\n'; return 0;} |
Java
// Java program to find// the smallest subarray having// atleast one duplicateimport java.util.*;class GFG { // Function to calculate // SubArray Length static int subArrayLength(int arr[], int n) { int minLen = Integer.MAX_VALUE; // Last stores the index of the last // occurrence of the corresponding value HashMap<Integer, Integer> last = new HashMap<Integer, Integer>(); for (int i = 0; i < n; i++) { // If the element has already occurred if (last.containsKey(arr[i]) && last.get(arr[i]) != 0) { minLen = Math.min(minLen, i - last.get(arr[i]) + 2); } last.put(arr[i], i + 1); } if (minLen == Integer.MAX_VALUE) { return -1; } return minLen; } // Driver Code public static void main(String[] args) { int n = 7; int arr[] = { 1, 2, 3, 1, 5, 4, 5 }; int ans = subArrayLength(arr, n); System.out.print(ans); }}// This code is contributed by 29AjayKumar |
Python
# Python program for above approachn = 7arr = [1, 2, 3, 1, 5, 4, 5]last = dict()minLen = n + 1for i in range(0, n): if arr[i] in last: minLen = min(minLen, i-last[arr[i]]+2) last[arr[i]]= i + 1 if minLen == n + 1: print("-1")else: print(minLen) |
C#
// C# program to find// the smallest subarray having// atleast one duplicateusing System;using System.Collections.Generic;class GFG { // Function to calculate // SubArray Length static int subArrayLength(int []arr, int n) { int minLen = int.MaxValue; // Last stores the index of the last // occurrence of the corresponding value Dictionary<int, int> last = new Dictionary<int, int>(); for (int i = 0; i < n; i++) { // If the element has already occurred if (last.ContainsKey(arr[i]) && last[arr[i]] != 0) { minLen = Math.Min(minLen, i - last[arr[i]] + 2); } if(last.ContainsKey(arr[i])) last[arr[i]] = i + 1; else last.Add(arr[i], i + 1); } if (minLen == int.MaxValue) { return -1; } return minLen; } // Driver Code public static void Main(String[] args) { int n = 7; int []arr = { 1, 2, 3, 1, 5, 4, 5 }; int ans = subArrayLength(arr, n); Console.Write(ans); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript program to find// the smallest subarray having// atleast one duplicate// Function to calculate // SubArray Length function subArrayLength(arr, n) { let minLen = Number.MAX_VALUE; // Last stores the index of the last // occurrence of the corresponding value let last = new Map(); for (let i = 0; i < n; i++) { // If the element has already occurred if (last.has(arr[i]) && last.get(arr[i]) != 0) { minLen = Math.min(minLen, i - last.get(arr[i]) + 2); } last.set(arr[i], i + 1); } if (minLen == Number.MAX_VALUE) { return -1; } return minLen; }// Driver code let n = 7; let arr = [ 1, 2, 3, 1, 5, 4, 5 ]; let ans = subArrayLength(arr, n); document.write(ans); </script> |
Output:
3
Time Complexity: O(N), where N is size of array
Auxiliary Space: O(N) because it is using unordered_map last
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
