Previous Smaller Element

Given an array arr[], find the Previous Smaller Element (PSE) for every element in the array.

• The Previous Smaller Element of an element x is defined as the first element to its left in the array that is smaller than x.
• If no such element exists for a particular position, the PSE should be considered as -1.

 Examples: 

Input: arr[] = [1, 6, 2]
Output: [-1, 1, 1]
Explanation: For the first element 1, there is no element to its left, so the result is -1. For 6, the previous smaller element is 1. For 2, the previous smaller element is also 1, since it is the closest smaller number when looking left.

Input: arr[] = [1, 5, 0, 3, 4, 5]
Output: [-1, 1, -1, 0, 3, 4]
Explanation:
For 1, no element on the left → -1
For 5, the previous smaller element is 1
For 0, no smaller element on the left → -1
For 3, the previous smaller element is 0
For 4, the previous smaller element is 3
For the last 5, the previous smaller element is 4

[Naive Approach] Using Nested Loops - O(n²) Time and O(1) Space

The idea is to use two loops: for each element, check the elements on its left to find the previous smaller one. If none exists, store -1.

#include <iostream>
#include <vector>
using namespace std;

vector<int> prevSmaller(vector<int>& arr){
    int n = arr.size();
    vector<int> result(n, -1);

    // for each element, check all elements 
    // on the left
    for (int i = 0; i < n; i++) {
        for (int j = i - 1; j >= 0; j--) {
            if (arr[j] < arr[i]) {
                result[i] = arr[j];
                break;
            }
        }
    }
    return result;
}

int main() {
    vector<int> arr = {1, 5, 0, 3, 4, 5};
    vector<int> ans = prevSmaller(arr);

    for (int x : ans) cout << x << " ";
    return 0;
}

#include <stdio.h>
#include <stdlib.h>

int* prevSmaller(int arr[], int n) {
    
    // allocate memory for result array
    int* result = (int*)malloc(n * sizeof(int));

    // initialize all PSEs as -1
    for (int i = 0; i < n; i++) result[i] = -1;

    for (int i = 0; i < n; i++) {
       
        // check all elements on the left
        for (int j = i - 1; j >= 0; j--) {
            if (arr[j] < arr[i]) {
                
                // first smaller element on the left
                result[i] = arr[j];
                break;
            }
        }
    }
    return result;
}

int main() {
    int arr[] = {1, 5, 0, 3, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);

    int* result = prevSmaller(arr, n);

    for (int i = 0; i < n; i++) {
        printf("%d ", result[i]);
    }
    printf("\n");

    return 0;
}

import java.util.ArrayList;

class GfG {
    static ArrayList<Integer> prevSmaller(int arr[]) {
        int n = arr.length;
        ArrayList<Integer> result = new ArrayList<>();

        // initialize all as -1
        for (int i = 0; i < n; i++) result.add(-1);

        // for each element, check all elements 
        // on the left
        for (int i = 0; i < n; i++) {
            for (int j = i - 1; j >= 0; j--) {
                if (arr[j] < arr[i]) {
                    result.