Given an array **arr[]** of size **N**, the task is for each array element is to find the nearest non-equal value present on its left in the array. If no such element is found, then print **-1**

**Examples:**

Input:arr[] = { 2, 1, 5, 8, 3 }Output:-1 2 2 5 2Explanation:

[2], it is the only number in this prefix. Hence, answer is -1.

[2,1], the closest number to 1 is 2

[2, 1,5], the closest number to 5 is 2

[2, 1, 5,8], the closest number to 8 is 5

[2, 1, 5, 8,3], the closest number to 3 is 2

Input:arr[] = {3, 3, 2, 4, 6, 5, 5, 1}Output:-1 -1 3 3 4 4 4 2Explanation:

[3], it is the only number in this prefix. Hence, answer is -1.

[3, 3], it is the only number in this prefix. Hence, answer is -1

[3, 3, 2], the closest number to 2 is 3

[3, 3, 2, 4], the closest number to 4 is 3

[3, 3, 2, 4, 6], the closest number to 6 is 4

[3, 3, 2, 4, 6, 5], the closest number to 5 is 4

[3, 3, 2, 4, 6, 5, 5], the closest number to 5 is 4

[3, 3, 2, 4, 6, 5, 5, 1], the closest number to 1 is 2

**Naive Approach:** The simplest idea is to traverse the given array and for every **i ^{th}** element, find the closest element on the left side of index

**i**which is not equal to

**arr[i]**.

**Time Complexity:**O(N^2)

**Auxiliary Space:**O(1)**Efficient Approach: **

The idea is to insert the elements of the given array in a Set such that the inserted numbers are sorted and then for an integer, find its position and compare its next value with the previous value, and print the closer value out of the two.

Fullow the steps below to sulve the problem:

- Initialize a Set of integers
**S**to store the elements in sorted order. - Traverse the array
**arr[]**using the variable**i**. - Now, find the nearest value smaller as well as greater than
**arr[i]**, say**X**and**Y**respectively.- If
**X**cannot be found, print**Y**. - If
**Y**cannot be found, print**Z**. - If both
**X**and**Y**cannot be found, print**“-1”**.

- If
- After that, add
**arr[i]**to the Set**S**and print**X**if**abs(X – arr[i])**is smaller than**abs(Y – arr[i])**. Otherwise, print**Y**. - Repeat the above steps for every element.

Below is the implementation of the above approach:

## C++14

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the closest number on` `// the left side of x` `void` `printClosest(set<` `int` `>& streamNumbers, ` `int` `x)` `{` ` ` `// Insert the value in set and store` ` ` `// its position` ` ` `auto` `it = streamNumbers.insert(x).first;` ` ` `// If x is the smallest element in set` ` ` `if` `(it == streamNumbers.begin()) ` ` ` `{` ` ` `// If count of elements in the set` ` ` `// equal to 1` ` ` `if` `(next(it) == streamNumbers.end()) ` ` ` `{` ` ` `cout << ` `"-1 "` `;` ` ` `return` `;` ` ` `}` ` ` `// Otherwise, print its` ` ` `// immediate greater element` ` ` `int` `rightVal = *next(it);` ` ` `cout << rightVal << ` `" "` `;` ` ` `return` `;` ` ` `}` ` ` `// Store its immediate smaller element` ` ` `int` `leftVal = *prev(it);` ` ` `// If immediate greater element does not` ` ` `// exists print it's immediate` ` ` `// smaller element` ` ` `if` `(next(it) == streamNumbers.end()) {` ` ` `cout << leftVal << ` `" "` `;` ` ` `return` `;` ` ` `}` ` ` `// Store the immediate` ` ` `// greater element` ` ` `int` `rightVal = *next(it);` ` ` `// Print the closest number` ` ` `if` `(x - leftVal <= rightVal - x)` ` ` `cout << leftVal << ` `" "` `;` ` ` `else` ` ` `cout << rightVal << ` `" "` `;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array` ` ` `vector<` `int` `> arr = { 3, 3, 2, 4, 6, 5, 5, 1 };` ` ` `// Initialize set` ` ` `set<` `int` `> streamNumbers;` ` ` `// Print Answer` ` ` `for` `(` `int` `i = 0; i < arr.size(); i++) {` ` ` `// Function Call` ` ` `printClosest(streamNumbers, arr[i]);` ` ` `}` ` ` `return` `0;` `}` |

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**Output**

-1 -1 3 3 4 4 4 2

**Time Complexity: **O(N * log(N))**Auxiliary Space: **O(N)

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