Print next greater number of Q queries

30 Jun, 2022
Given an array of n elements and q queries, for each query that has index i, find the next greater element and print its value. If there is no such greater element to its right then print -1. Examples:  
Input : arr[] = {3, 4, 2, 7, 5, 8, 10, 6} 
        query indexes = {3, 6, 1}
Output: 8 -1 7 
Explanation : 
For the 1st query index is 3, element is 7 and 
the next greater element at its right is 8 

For the 2nd query index is 6, element is 10 and 
there is no element greater than 10 at right, 
so print -1.

For the 3rd query index is 1, element is 4 and
the next greater element at its right is 7.
 
Normal Approach: A normal approach will be for every query to move in a loop from index to n and find out the next greater element and print it, but this in worst case will take n iterations, which is a lot if the number of queries are high. Time Complexity: O(n^2) Auxiliary Space>: O(1)Efficient Approach: An efficient approach is based on next greater element. We store the index of the next greater element in an array and for every query process, answer the query in O(1) that will make it more efficient. But to find out the next greater element for every index in array there are two ways. One will take o(n^2) and O(n) space which will be to iterate from I+1 to n for each element at index I and find out the next greater element and store it.But the more efficient one will be to use stack, where we use indexes to compare and store in next[] the next greater element index.1) Push the first index to stack. 2) Pick rest of the indexes one by one and follow following steps in loop. ....a) Mark the current element as i. ....b) If stack is not empty, then pop an index from stack and compare a[index] with a[I]. ....c) If a[I] is greater than the a[index], then a[I] is the next greater element for the a[index]. ....d) Keep popping from the stack while the popped index element is smaller than a[I]. a[I] becomes the next greater element for all such popped elements ....g) If a[I] is smaller than the popped index element, then push the popped index back.3) After the loop in step 2 is over, pop all the index from stack and print -1 as next index for them. 
C++
Java
Python3
C#
Javascript
// C++ program to print// next greater number// of Q queries#include <bits/stdc++.h>using namespace std; // array to store the next// greater element indexvoid next_greatest(int next[],                   int a[], int n){    // use of stl    // stack in c++    stack<int> s;     // push the 0th    // index to the stack    s.push(0);     // traverse in the    // loop from 1-nth index    for (int i = 1; i < n; i++)    {         // iterate till loop is empty        while (!s.empty()) {             // get the topmost            // index in the stack            int cur = s.top();             // if the current element is             // greater than the top indexth            // element, then this will be            // the next greatest index            // of the top indexth element            if (a[cur] < a[i])            {                                 // initialise the cur                // index position's                // next greatest as index                next[cur] = i;                 // pop the cur index                // as its greater                // element has been found                s.pop();            }             // if not greater            // then break            else                break;        }        // push the i index so that its        // next greatest can be found        s.push(i);    }     // iterate for all other    // index left inside stack    while (!s.empty())    {        int cur = s.top();         // mark it as -1 as no        // element in greater        // then it in right        next[cur] = -1;         s.pop();    }} // answers all// queries in O(1)int answer_query(int a[], int next[],                 int n, int index){    // stores the next greater    // element positions    int position = next[index];     // if position is -1 then no    // greater element is at right.    if (position == -1)        return -1;     // if there is a index that    // has greater element    // at right then return its    // value as a[position]    else        return a[position];} // Driver Codeint main(){     int a[] = {3, 4, 2, 7,               5, 8, 10, 6 };     int n = sizeof(a) / sizeof(a[0]);     // initializes the    // next array as 0    int next[n] = { 0 };     // calls the function    // to pre-calculate    // the next greatest    // element indexes    next_greatest(next, a, n);     // query 1 answered    cout << answer_query(a, next, n, 3) << " ";     // query 2 answered    cout << answer_query(a, next, n, 6) << " ";     // query 3 answered    cout << answer_query(a, next, n, 1) << " ";}
// Java program to print// next greater number// of Q queriesimport java.util.*; class GFG{public static int[] query(int arr[],                          int query[]){    int ans[] = new int[arr.length];// this array contains                                    // the next greatest                                    // elements of all the elements    Stack<Integer> s = new Stack<>();    // push the 0th index    // to the stack    s.push(arr[0]);    int j = 0;    //traverse rest    // of the array    for(int i = 1; i < arr.length; i++)    {        int next = arr[i];                 if(!s.isEmpty())        {            // get the topmost            // element in the stack            int element = s.pop();                         /* If the popped element            is smaller than next,            then a) store the pair            b) keep popping while            elements are smaller and            stack is not empty */            while(next > element)            {                ans[j] = next;                j++;                if(s.isEmpty())                    break;                element = s.pop();            }                         /* If element is greater            than next, then            push the element back */            if (element > next)                s.push(element);        }        /* push next to stack so        that we can find next        greater for it */        s.push(next);    }    /* After iterating over the    loop, the remaining elements    in stack do not have the next    greater element, so -1 for them */    while(!s.isEmpty())    {        int element = s.pop();        ans[j] = -1;        j++;    }         // return the next    // greatest array    return ans;} // Driver Code   public static void main(String[] args){    int arr[] = {3, 4, 2, 7,                 5, 8, 10, 6};    int query[] = {3, 6, 1};    int ans[] = query(arr,query);         // getting output array    // with next greatest elements    for(int i = 0; i < query.length; i++)    {        // displaying the next greater        // element for given set of queries        System.out.print(ans[query[i]] + " ");    }}} // This code was contributed// by Harshit Sood
# Python3 program to print# next greater number# of Q queries # array to store the next# greater element indexdef next_greatest(next, a, n):     # use of stl    # stack in c++    s = []     # push the 0th    # index to the stack    s.append(0);     # traverse in the    # loop from 1-nth index    for  i in range(1, n):         # iterate till loop is empty        while (len(s) != 0):             # get the topmost            # index in the stack            cur = s[-1]             # if the current element is             # greater then the top indexth            # element, then this will be            # the next greatest index            # of the top indexth element            if (a[cur] < a[i]):                                 # initialise the cur                # index position's                # next greatest as index                next[cur] = i;                 # pop the cur index                # as its greater                # element has been found                s.pop();             # if not greater            # then break            else:                break;                 # push the i index so that its        # next greatest can be found        s.append(i);     # iterate for all other    # index left inside stack    while(len(s) != 0):        cur = s[-1]         # mark it as -1 as no        # element in greater        # then it in right        next[cur] = -1;        s.pop();     # answers all# queries in O(1)def answer_query(a, next, n, index):     # stores the next greater    # element positions    position = next[index];     # if position is -1 then no    # greater element is at right.    if(position == -1):        return -1;     # if there is a index that    # has greater element    # at right then return its    # value as a[position]    else:        return a[position]; # Driver Codeif __name__=='__main__':     a = [3, 4, 2, 7, 5, 8, 10, 6 ]    n = len(a)     # initializes the    # next array as 0    next=[0 for i in range(n)]     # calls the function    # to pre-calculate    # the next greatest    # element indexes    next_greatest(next, a, n);     # query 1 answered    print(answer_query(a, next, n, 3), end = ' ')     # query 2 answered    print(answer_query(a, next, n, 6), end = ' ')     # query 3 answered    print(answer_query(a, next, n, 1), end = ' ') # This code is contributed by rutvik_56.
// C# program to print next greater// number of Q queriesusing System;using System.Collections.Generic; class GFG{public static int[] query(int[] arr,                          int[] query){    int[] ans = new int[arr.Length]; // this array contains                                     // the next greatest                                     // elements of all the elements    Stack<int> s = new Stack<int>();         // push the 0th index to the stack    s.Push(arr[0]);    int j = 0;         // traverse rest of the array    for (int i = 1; i < arr.Length; i++)    {        int next = arr[i];         if (s.Count > 0)        {            // get the topmost element in the stack            int element = s.Pop();             /* If the popped element is smaller            than next, then            a) store the pair            b) keep popping while            elements are smaller and            stack is not empty */            while (next > element)            {                ans[j] = next;                j++;                if (s.Count == 0)                {                    break;                }                element = s.Pop();            }             /* If element is greater            than next, then            push the element back */            if (element > next)            {                s.Push(element);            }        }        /* push next to stack so that we        can find next greater for it */        s.Push(next);    }         /* After iterating over the    loop, the remaining elements    in stack do not have the next    greater element, so -1 for them */    while (s.Count > 0)    {        int element = s.Pop();        ans[j] = -1;        j++;    }     // return the next greatest array    return ans;} // Driver Code    public static void Main(string[] args){    int[] arr = new int[] {3, 4, 2, 7, 5, 8, 10, 6};    int[] query = new int[] {3, 6, 1};    int[] ans = GFG.query(arr, query);     // getting output array    // with next greatest elements    for (int i = 0; i < query.Length; i++)    {        // displaying the next greater        // element for given set of queries        Console.Write(ans[query[i]] + " ");    }}} // This code is contributed by Shrikant13
<script> // JavaScript program to print// next greater number// of Q queries // array to store the next// greater element indexfunction next_greatest(next, a, n){    // use of stl    // stack in c++    var s = [];     // push the 0th    // index to the stack    s.push(0);     // traverse in the    // loop from 1-nth index    for (var i = 1; i < n; i++)    {         // iterate till loop is empty        while (s.length!=0) {             // get the topmost            // index in the stack            var cur = s[s.length-1];             // if the current element is             // greater then the top indexth            // element, then this will be            // the next greatest index            // of the top indexth element            if (a[cur] < a[i])            {                                 // initialise the cur                // index position's                // next greatest as index                next[cur] = i;                 // pop the cur index                // as its greater                // element has been found                s.pop();            }             // if not greater            // then break            else                break;        }        // push the i index so that its        // next greatest can be found        s.push(i);    }     // iterate for all other    // index left inside stack    while (s.length!=0)    {        var cur = s[s.length-1];         // mark it as -1 as no        // element in greater        // then it in right        next[cur] = -1;         s.pop();    }} // answers all// queries in O(1)function answer_query(a, next, n, index){    // stores the next greater    // element positions    var position = next[index];     // if position is -1 then no    // greater element is at right.    if (position == -1)        return -1;     // if there is a index that    // has greater element    // at right then return its    // value as a[position]    else        return a[position];} // Driver Codevar a = [3, 4, 2, 7,           5, 8, 10, 6];var n = a.length;// initializes the// next array as 0var next = Array(n).fill(0);// calls the function// to pre-calculate// the next greatest// element indexesnext_greatest(next, a, n);// query 1 answereddocument.write( answer_query(a, next, n, 3) + " ");// query 2 answereddocument.write( answer_query(a, next, n, 6) + " ");// query 3 answereddocument.write( answer_query(a, next, n, 1) + " ");   </script>
Output:  
8 -1 7 
Time complexity: max(O(n), O(q)), O(n) for pre-processing the next[] array and O(1) for every query.Auxiliary Space: O(n)This article is contributed by Raja Vikramaditya(raj). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
HarshitSood1
shrikanth13
rutvik_56
itsok
simmytarika5
mitalibhola94
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