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Stack_NextLargerElement.java
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Stack_NextLargerElement.java
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import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Stack;
// https://practice.geeksforgeeks.org/problems/next-larger-element/0
// https://www.youtube.com/watch?v=8P-Z7Oc8x9I
// https://www.youtube.com/watch?v=sDKpIO2HGq0 (store the indices to print in same order as input)
// https://www.youtube.com/watch?v=R2o2zUuu3ss
// https://www.geeksforgeeks.org/next-greater-element-in-same-order-as-input/
class Stack_NextLargerElement {
// The benefit is that when we arrive at a certain index his next greater element will be already in stack
// and we can directly get this element while at the same index.
private static long[] findNextGreaterElement(long[] ip, int n) {
long[] op = new long[n];
Stack<Long> stack = new Stack<>();
for(int i = n-1; i >= 0; i--) {
// keep popping until the top of stack is not greater than the current value
while(!stack.isEmpty() && stack.peek() <= ip[i]) {
stack.pop();
}
op[i] = (stack.isEmpty()) ? -1 : stack.peek();
stack.push(ip[i]);
}
stack.clear();
return op;
}
public static void main(String[] args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-->0)
{
int n = Integer.parseInt(br.readLine());
String[] str = br.readLine().trim().split("\\s+");
long[] ip = new long[n];
for(int i = 0; i < n; i++) {
ip[i] = Long.parseLong(str[i]);
}
// Approach-3; this approach print the output in the same order as input
long[] op = findNextGreaterElement(ip,n);
print(op, n);
}
}
private static void print(long[] ip, int n) {
StringBuffer sb = new StringBuffer();
for(int i = 0; i < n; i++) {
sb.append(ip[i] + " ");
}
System.out.println(sb);
}
}
// -------------------------------------------------------------------------------------------
// https://www.youtube.com/watch?v=sDKpIO2HGq0 (store the indices to print in same order as input)
/*
// in stack, we will store the indices of the actual input.
private static long[] findNextGreaterElement(long[] ip, int n) {
long[] nge = new long[n];
Stack<Integer> stack = new Stack<>();
stack.push(0);
for(int i = 1; i < n; i++) {
long num = ip[i];
while(!stack.isEmpty() && ip[stack.peek()] < num ) {
nge[stack.pop()] = num;
}
stack.push(i);
}
while(!stack.isEmpty()) {
nge[stack.pop()] = -1;
}
return nge;
}
private static void print(long[] ip, int n) {
StringBuffer sb = new StringBuffer();
for(int i = 0; i < n; i++) {
sb.append(ip[i] + " ");
}
System.out.println(sb);
}
*/
// -------------------------------------------------------------------------------------------
// Approach-2; this approach doesn't print the output in the same order as input
/*Stack<Integer> st = new Stack<>();
st.push(Integer.parseInt(ip[0])); //pushing the first element in the stack;
for (int i = 1; i < n; i++) {
int no = Integer.parseInt(ip[i]);
while( !st.isEmpty() && (st.peek() <= no) ) {
System.out.println("Num "+st.peek() +" NGE "+no);
st.pop();
}
st.push(no);
}
while(!st.isEmpty()) {
System.out.println("Num "+st.pop() +" NGE "+-1);
}*/
// -------------------------------------------------------------------------------------------
// Approach-1; 0(n^2)
/*
int a[] = new int[n];
for (int i = 0; i < ip.length-1; i++) {
int op = a[i];
for (int j = i+1; j < ip.length; j++) {
if(a[j] > op) {
op = a[j];
break;
}
}
if(op==a[i])
{
a[i] = -1;
}
else
{
a[i] = op;
}
}
a[n-1]= -1;
for (int i = 0; i < ip.length; i++) {
System.out.print(a[i]+" ");
}
System.out.println();*/