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
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;
}
}