02 First Repeating Element

First Repeating Element | Practice | GeeksforGeeks

Question

Given an array arr[] of size n, find the first repeating element. The element should occur more than once and the index of its first occurrence should be the smallest.

Example 1

Input:
n = 7
arr[] = {1, 5, 3, 4, 3, 5, 6}
Output: 2
Explanation: 
5 is appearing twice and 
its first appearence is at index 2 
which is less than 3 whose first 
occuring index is 3.

Example 2

Input:
n = 4
arr[] = {1, 2, 3, 4}
Output: -1
Explanation: 
All elements appear only once so 
the answer is -1.

The function firstRepeated() takes arr and n as input parameters and returns the position of the first repeating element. If there is no such element -1 is returned. 1-based indexing is used.

Expected Time Complexity: $O(n)$

Expected Auxilliary Space: $O(n)$

Constraints

$1 \le n \le 10^6$

$0 \le A_i \le 10^6$​

Naive Approach

Logic

1. Run two nested loops

2. For every element from the outer loop

   1. Check all the elements after that element

      1. If the same element is found then return the iterator of the outer loop

3. If still nothing found then return -1

Code

int firstRepeated(int arr[], int n) {
    for(int i = 0; i < n; i++)
        for(int j = i + 1; j < n; j++)
            if(arr[i] == arr[j])
                return i + 1;
                
    return -1;
}

Complexity

Time Complexity: $O(n^2)$

Space Complexity: $O(1)$​

Using map

Logic

1. firstRepeating = -1 stores the index of the first repeating element.

2. We traverse the array from right to left.

3. A set visited will store those elements that we have encountered on the right side of the current element.

4. While traversing from right to left

   1. If any element has already been visited

      1. It is a repeating element.

         1. Suppose this is the first occurence of the repeating element.

            1. Update firstRepeating

   2. Otherwise add it to the visited list

5. In the end, either we have -1 in firstRepeating which tells that there are no repetitive elements, or we find the index of the repeating element.

Code

int firstRepeated(int arr[], int n) {
    int firstRepeating = -1;
    
    unordered_set<int> visited;
    
    for(int i = n - 1; i >= 0; i--) {
        if(visited.count(arr[i]))
            firstRepeating = i;
        visited.insert(arr[i]);
    }
    
    return ((firstRepeating == -1) ? -1 : firstRepeating + 1);
}

Complexity

Time Complexity: $O(n)$​

Space Complexity: $O(n)$​

Using set

int firstRepeated(int arr[], int n) {
    int firstRepeating = -1;
    
    unordered_set<int> visited;
    
    for(int i = n - 1; i >= 0; i--) {
        if(visited.count(arr[i]))
            firstRepeating = i;
        visited.insert(arr[i]);
    }
    
    return ((firstRepeating == -1) ? -1 : firstRepeating + 1);
}