Print next greater number of Q queries
Last Updated :
23 Apr, 2025
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 the worst case will take n iterations, which is a lot if the number of queries are high.
Steps to implement-
- Run a loop for taking each query one by one
- Initialize a boolean variable as false to check whether the next greater element for any query exists or not
- Run a loop from the next index as given in the query to the last
- If we will get the next greater element then print that and make the boolean variable true, to indicate next greater element exists and break that inner loop
- After the inner loop, if the boolean variable is false then its next greater element doesn't exist. So print -1 for that
Code-
C++
// C++ program to print next
//greater number of Q queries
#include <bits/stdc++.h>
using namespace std;
// Function to print next greater
// of Q query
void next_greater(int arr[],int query[], int n,int q)
{
//Loop for traversing each query
for(int i=0;i<q;i++){
//Take every query one by one
int k=query[i];
//To detect next greater element is present or not
bool val=false;
//Loop for checking next greater of each query
for(int j=k+1;j<n;j++){
if(arr[j]>arr[k]){
cout<<arr[j]<<" ";
//Make val true when there exist a next greater element
val=true;
break;
}
}
//If next greater element is not present
if(val==false){
cout<<-1<<" ";
}
}
}
// Driver Code
int main()
{
//Given Numbers
int arr[] = {3, 4, 2, 7,5, 8, 10, 6 };
//Queries
int query[]={3,6,1};
int n = sizeof(arr) / sizeof(arr[0]);
int q= sizeof(query) / sizeof(query[0]);
next_greater(arr,query,n,q);
}
Java
import java.util.Arrays;
public class GFG {
// Function to print the next greater elements for Q queries
public static void nextGreater(int[] arr, int[] query) {
int n = arr.length;
int q = query.length;
// Loop for traversing each query
for (int i = 0; i < q; i++) {
// Take every query one by one
int k = query[i];
// To detect if the next greater element is present or not
boolean val = false;
// Loop for checking the next greater element for each query
for (int j = k + 1; j < n; j++) {
if (arr[j] > arr[k]) {
System.out.print(arr[j] + " ");
// Set val to true when there exists a next greater element
val = true;
break;
}
}
// If the next greater element is not present
if (!val) {
System.out.print(-1 + " ");
}
}
}
public static void main(String[] args) {
// Given numbers
int[] arr = { 3, 4, 2, 7, 5, 8, 10, 6 };
// Queries
int[] query = { 3, 6, 1 };
int n = arr.length;
int q = query.length;
nextGreater(arr, query);
}
}
Python
def next_greater_elements(arr, query):
# Initialize a list to store results as strings
results = []
# Loop through each query
for k in query:
# Initialize a flag to detect if the next greater element is found
val = False
# Loop to check for the next greater element
for j in range(k + 1, len(arr)):
if arr[j] > arr[k]:
results.append(str(arr[j]))
# Set the flag to True when a next greater element is found
val = True
break
# If no next greater element is found, push "-1" to the results list
if not val:
results.append("-1")
# Print the results as a space-separated string
print(" ".join(results))
# Given numbers
arr = [3, 4, 2, 7, 5, 8, 10, 6]
# Queries
query = [3, 6, 1]
next_greater_elements(arr, query)
C#
using System;
class GFG {
// Function to print next greater of Q queries
static void NextGreater(int[] arr, int[] query, int n,
int q)
{
// Loop for traversing each query
for (int i = 0; i < q; i++) {
// Take every query one by one
int k = query[i];
// To detect next greater element is present or
// not
bool val = false;
// Loop for checking next greater of each query
for (int j = k + 1; j < n; j++) {
if (arr[j] > arr[k]) {
Console.Write(arr[j] + " ");
// Make val true when there exists a
// next greater element
val = true;
break;
}
}
// If next greater element is not present
if (val == false) {
Console.Write(-1 + " ");
}
}
}
// Driver Code
static void Main(string[] args)
{
// Given Numbers
int[] arr = { 3, 4, 2, 7, 5, 8, 10, 6 };
// Queries
int[] query = { 3, 6, 1 };
int n = arr.Length;
int q = query.Length;
NextGreater(arr, query, n, q);
}
}
JavaScript
function GFG(arr, query) {
// Initialize an array to store results as a string
const results = [];
// Loop through each query
for (let i = 0; i < query.length; i++) {
// Take each query one by one
const k = query[i];
// Initialize a flag to detect if the next
// greater element is found
let val = false;
// Loop to check for the next greater element
for (let j = k + 1; j < arr.length; j++) {
if (arr[j] > arr[k]) {
results.push(arr[j].toString());
// Set the flag to true when a next
// greater element is found
val = true;
break;
}
}
// If no next greater element is found
// push "-1" to the results array
if (!val) {
results.push("-1");
}
}
// Print the results as a space-separated string
console.log(results.join(" "));
}
// Given numbers
const arr = [3, 4, 2, 7, 5, 8, 10, 6];
// Queries
const query = [3, 6, 1];
GFG(arr, query);
Time Complexity: O(n^2) ,because of two nested loops
Auxiliary Space>: O(1) , because no extra space has been used
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++
// C++ program to print
// next greater number
// of Q queries
#include <bits/stdc++.h>
using namespace std;
// array to store the next
// greater element index
void 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 Code
int 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
// Java program to print
// next greater number
// of Q queries
import java.util.*;
class GFG
{
public static int[] findGreaterElements(int arr[]) {
int ans[] = new int[arr.length];
Stack<Integer> s = new Stack<>();
s.push(arr[arr.length - 1]);
ans[arr.length - 1] = -1;
for (int i = arr.length - 2; i >= 0; i--) {
int curr = arr[i];
if (s.isEmpty()) {
ans[i] = -1;
} else {
if (s.peek() <= curr) {
while (s.peek() <= curr) {
s.pop();
if (s.isEmpty()) {
break;
}
}
if (s.isEmpty()) {
ans[i] = -1;
} else {
ans[i] = s.peek();
}
} else {
ans[i] = s.peek();
}
s.push(curr);
}
}
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[] = findGreaterElements(arr);
// 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 Raj Suvariya
Python
# Python3 program to print
# next greater number
# of Q queries
# array to store the next
# greater element index
def 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 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.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 Code
if __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#
// C# program to print next greater
// number of Q queries
using 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
JavaScript
// JavaScript program to print
// next greater number
// of Q queries
// array to store the next
// greater element index
function 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 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.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 Code
var a = [3, 4, 2, 7,
5, 8, 10, 6];
var n = a.length;
// initializes the
// next array as 0
var next = Array(n).fill(0);
// calls the function
// to pre-calculate
// the next greatest
// element indexes
next_greatest(next, a, n);
// query 1 answered
console.log( answer_query(a, next, n, 3) + " ");
// query 2 answered
console.log( answer_query(a, next, n, 6) + " ");
// query 3 answered
console.log( answer_query(a, next, n, 1) + " ");
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)
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