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# Nearest Smaller Element

Given an array, find the **nearest** smaller element G\[i] for every element A\[i] in the array such that the element has an **index smaller than i**.

More formally,

```
    G[i] for an element A[i] = an element A[j] such that 
    j is maximum possible AND 
    j < i AND
    A[j] < A[i]
```

Elements for which no smaller element exist, consider next smaller element as -1.

**Input Format**

```
The only argument given is integer array A.
```

**Output Format**

```
Return the integar array G such that G[i] contains nearest smaller number than A[i].If no such element occurs G[i] should be -1.
```

**For Example**

```
Input 1:
    A = [4, 5, 2, 10, 8]
Output 1:
    G = [-1, 4, -1, 2, 2]
Explaination 1:
    index 1: No element less than 4 in left of 4, G[1] = -1
    index 2: A[1] is only element less than A[2], G[2] = A[1]
    index 3: No element less than 2 in left of 2, G[3] = -1
    index 4: A[3] is nearest element which is less than A[4], G[4] = A[3]
    index 4: A[3] is nearest element which is less than A[5], G[5] = A[3]
    
Input 2:
    A = [3, 2, 1]
Output 2:
    [-1, -1, -1]
Explaination 2:
    index 1: No element less than 3 in left of 3, G[1] = -1
    index 2: No element less than 2 in left of 2, G[2] = -1
    index 3: No element less than 1 in left of 1, G[3] = -1
```

```java
public class Solution {
    public int[] prevSmaller(int[] A) {
        Stack<Integer> st = new Stack<>();
        int[] ans = new int[A.length];
        for (int i = 0; i < A.length; i++) {
            if (st.size() == 0) {
                st.push(A[i]);
                ans[i] = -1;
            } else {
                if (st.peek() >= A[i]) {
                    while (st.size() != 0 && st.peek() >= A[i])
                        st.pop();
                    if (st.size() == 0)
                        ans[i] = -1;
                    else
                        ans[i] = st.peek();
                    st.push(A[i]);
                } else {
                    ans[i] = st.peek();
                    st.push(A[i]);
                }
            }
        }
        return ans;
    }
}
```