79 lines
2.3 KiB
Java
79 lines
2.3 KiB
Java
/***
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* Problem Statement:
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* Given an array arr[] of size N having distinct elements, the task is to find the next greater element for each element of the array in order of their appearance in the array.
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* Next greater element means nearest element on right side which is greater than current element,
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* if there is no suxh element, then put -1 for that.
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*
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*/
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/***
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* Example 1: Input: arr=[1,3,2,4] n=4
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Output: 3 4 4 -1
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Example 2: Input: arr=[6 8 0 1 3] n=5
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Output: 8 -1 1 3 -1
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*/
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/***
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* Time Complexity: O(N)
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* Space Complexity: O(N)
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*/
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import java.util.Scanner;
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import java.util.Stack;
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public class Next_Greater_Element {
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public static void main(String[] args){
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Scanner input=new Scanner(System.in);
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int n=input.nextInt();
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long arr[]=new long[n];
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for (int i=0;i<n;i++){
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arr[i]=input.nextLong();
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}
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long ans[]=nextLargerElement(arr,n); // functions that print array for calculating next greater element
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for (int i=0;i<ans.length;i++){
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System.out.print(ans[i]+" ")
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}
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}
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public static long[] nextLargerElement(long[] arr, int n)
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{
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// Your code here
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/****
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* Approach: for element at index i,
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* there were two posiibilities that there is greater element than this
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* at right side of array or not.
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* if there is no greater element in right side is indicated by Empty stack
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* if there is greater ekement in right side is indicated by stack peek element.
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*/
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long ans[]=new long[n];
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ans[n-1]=-1;
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Stack<Long> st=new Stack<>();
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st.push(arr[n-1]);
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for (int i=n-2;i>=0;i--)
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{
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if(st.isEmpty())
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{
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st.push(arr[i]);
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ans[i]=-1;
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}
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else
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{
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while(st.isEmpty()==false && arr[i]>=st.peek())
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{
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st.pop();
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}
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if(st.size()==0){
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ans[i]=-1;
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}else{
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ans[i]=st.peek();
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}
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st.push(arr[i]);
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}
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}
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return ans;
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}
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}
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