d

```
package com.thealgorithms.stacks;
import java.util.Arrays;
import java.util.Stack;
/*
Given an array "input" you need to print the first smaller element for each element to the left
side of an array. For a given element x of an array, the Next Smaller element of that element is
the first smaller element to the left side of it. If no such element is present print -1.
Example
input = { 2, 7, 3, 5, 4, 6, 8 };
At i = 0
No elements to left of it : -1
At i = 1
Next smaller element between (0 , 0) is 2
At i = 2
Next smaller element between (0 , 1) is 2
At i = 3
Next smaller element between (0 , 2) is 3
At i = 4
Next smaller element between (0 , 3) is 3
At i = 5
Next smaller element between (0 , 4) is 4
At i = 6
Next smaller element between (0 , 5) is 6
result : [-1, 2, 2, 3, 3, 4, 6]
1) Create a new empty stack st
2) Iterate over array "input" , where "i" goes from 0 to input.length -1.
a) We are looking for value just smaller than `input[i]`. So keep popping from "stack"
till elements in "stack.peek() >= input[i]" or stack becomes empty.
b) If the stack is non-empty, then the top element is our previous element. Else the
previous element does not exist. c) push input[i] in stack. 3) If elements are left then their
answer is -1
*/
public final class NextSmallerElement {
private NextSmallerElement() {
}
public static int[] findNextSmallerElements(int[] array) {
// base case
if (array == null) {
return array;
}
Stack<Integer> stack = new Stack<>();
int[] result = new int[array.length];
Arrays.fill(result, -1);
for (int i = 0; i < array.length; i++) {
while (!stack.empty() && stack.peek() >= array[i]) stack.pop();
if (stack.empty()) {
result[i] = -1;
} else {
result[i] = stack.peek();
}
stack.push(array[i]);
}
return result;
}
public static void main(String[] args) {
int[] input = {2, 7, 3, 5, 4, 6, 8};
int[] result = findNextSmallerElements(input);
System.out.println(Arrays.toString(result));
}
}
```