Given an array of n integers and q queries, print the number of next greater elements to the right of the given index element.
Examples:
Input: a[] = {3, 4, 2, 7, 5, 8, 10, 6}
q = 2
index = 0,
index = 5
Output: 6
1
Explanation:
The next greater elements
to the right of 3(index 0) are 4,7,5,8,10,6.
The next greater elements to the right
of 8(index 5) are 10.
A naive approach is to iterate for every query from index to end, and find out the number of next greater elements to the right. This wonât be efficient enough as we run two nested loops
Time Complexity: O(N) to answer a query.
Auxiliary space: O(1)
Better approach is to store the next greater index of every element and run a loop for every query that iterates from index and keeping the increasing counter as j = next[i]. This will avoid checking all elements and will directly jump to the next greater element of every element. But this wonât be efficient enough in cases like 1 2 3 4 5 6, where the next greater elements are sequentially increasing, ending it up in taking O(n) for every query.
Time complexity: O(N) to answer a query.
Auxiliary space: O(N) for next greater element.
Efficient approach is to store the next greater elements index using next greater element in a next[] array. Then create a dp[] array that starts from n-2, as n-1th index will have no elements to its right and dp[n-1] = 0. While traversing from back we use dynamic programming to count the number of elements to the right where we use memoization as dp[next[i]] which gives us a count of the numbers to the right of the next greater element of the current element, hence we add 1 to it. If next[i]=-1 then we do not have any element to the right hence dp[i]=0. dp[index] stores the count of the number of next greater elements to the right.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>using namespace std;// array to store the next greater element indexvoid fillNext(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 // then the top index-th element, then // this will be the next greatest index // of the top index-th element if (a[cur] < a[i]) { // initialize 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(); }}// Function to count the number of // next greater numbers to the rightvoid count(int a[], int dp[], int n){ // initializes the next array as 0 int next[n]; memset(next, 0, sizeof(next)); // calls the function to pre-calculate // the next greatest element indexes fillNext(next, a, n); for (int i = n - 2; i >= 0; i--) { // if the i-th element has no next // greater element to right if (next[i] == -1) dp[i] = 0; // Count of next greater numbers to right. else dp[i] = 1 + dp[next[i]]; }}// answers all queries in O(1)int answerQuery(int dp[], int index){ // returns the number of next greater // elements to the right. return dp[index];}// driver program to test the above functionint main(){ int a[] = { 3, 4, 2, 7, 5, 8, 10, 6 }; int n = sizeof(a) / sizeof(a[0]); int dp[n]; // calls the function to count the number // of greater elements to the right for // every element. count(a, dp, n); // query 1 answered cout << answerQuery(dp, 3) << endl; // query 2 answered cout << answerQuery(dp, 6) << endl; // query 3 answered cout << answerQuery(dp, 1) << endl; return 0;} |
Java
// Java program to print number of NGEs to the rightimport java.util.*;class GFG{// array to store the next greater element indexstatic void fillNext(int next[], int a[], int n){ // Use stack Stack<Integer> s = new Stack<Integer>(); // 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.size() > 0) { // get the topmost index in the stack int cur = s.peek(); // if the current element is greater // then the top index-th element, then // this will be the next greatest index // of the top index-th element if (a[cur] < a[i]) { // initialize 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.size() > 0) { int cur = s.peek(); // mark it as -1 as no element in greater // then it in right next[cur] = -1; s.pop(); }}// function to count the number of // next greater numbers to the rightstatic void count(int a[], int dp[], int n){ // initializes the next array as 0 int next[] = new int[n]; for(int i = 0; i < n; i++) next[i] = 0; // calls the function to pre-calculate // the next greatest element indexes fillNext(next, a, n); for (int i = n - 2; i >= 0; i--) { // if the i-th element has no next // greater element to right if (next[i] == -1) dp[i] = 0; // Count of next greater numbers to right. else dp[i] = 1 + dp[next[i]]; }}// answers all queries in O(1)static int answerQuery(int dp[], int index){ // returns the number of next greater // elements to the right. return dp[index];}// driver codepublic static void main(String args[]){ int a[] = { 3, 4, 2, 7, 5, 8, 10, 6 }; int n = a.length; int dp[] = new int[n]; // calls the function to count the number // of greater elements to the right for // every element. count(a, dp, n); // query 1 answered System.out.println(answerQuery(dp, 3)); // query 2 answered System.out.println( answerQuery(dp, 6)); // query 3 answered System.out.println( answerQuery(dp, 1) );}}// This code is contributed by Arnab Kundu |
C#
// C# program to print number of // NGEs to the rightusing System;using System.Collections;class GFG{// array to store the next greater element indexstatic void fillNext(int []next, int []a, int n){ // Use stack Stack s = new Stack(); // 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.Count > 0) { // get the topmost index in the stack int cur = (int)s.Peek(); // if the current element is greater // then the top index-th element, then // this will be the next greatest index // of the top index-th element if (a[cur] < a[i]) { // initialize 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.Count > 0) { int cur =(int) s.Peek(); // mark it as -1 as no element in // greater then it in right next[cur] = -1; s.Pop(); }}// function to count the number of // next greater numbers to the rightstatic void count(int []a, int []dp, int n){ // initializes the next array as 0 int []next = new int[n]; for(int i = 0; i < n; i++) next[i] = 0; // calls the function to pre-calculate // the next greatest element indexes fillNext(next, a, n); for (int i = n - 2; i >= 0; i--) { // if the i-th element has no next // greater element to right if (next[i] == -1) dp[i] = 0; // Count of next greater numbers to right. else dp[i] = 1 + dp[next[i]]; }}// answers all queries in O(1)static int answerQuery(int []dp, int index){ // returns the number of next greater // elements to the right. return dp[index];}// Driver codepublic static void Main(String []args){ int []a = { 3, 4, 2, 7, 5, 8, 10, 6 }; int n = a.Length; int []dp = new int[n]; // calls the function to count the number // of greater elements to the right for // every element. count(a, dp, n); // query 1 answered Console.WriteLine(answerQuery(dp, 3)); // query 2 answered Console.WriteLine( answerQuery(dp, 6)); // query 3 answered Console.WriteLine( answerQuery(dp, 1));}}// This code is contributed by Arnab Kundu |
2 0 3
Time complexity: O(1) to answer a query.
Auxiliary Space: O(n)
Alternate Shorter Implementation
C++
#include <bits/stdc++.h>using namespace std;vector<int> no_NGN(int arr[], int n){ vector<int> nxt; // use of stl stack in c++ stack<int> s; nxt.push_back(0); // push the (n-1)th index to the stack s.push(n - 1); // traverse in reverse order for (int i = n - 2; i >= 0; i--) { while (!s.empty() && arr[i] >= arr[s.top()]) s.pop(); // if no element is greater than arr[i] the // number of NGEs to right is 0 if (s.empty()) nxt.push_back(0); else // number of NGEs to right of arr[i] is // one greater than the number of NGEs to right // of higher number to its right nxt.push_back(nxt[n - s.top() - 1] + 1); s.push(i); } // reverse again because values are in reverse order reverse(nxt.begin(), nxt.end()); // returns the vector of number of next // greater elements to the right of each index. return nxt;}int main(){ int n = 8; int arr[] = { 3, 4, 2, 7, 5, 8, 10, 6 }; vector<int> nxt = no_NGN(arr, n); // query 1 answered cout << nxt[3] << endl; // query 2 answered cout << nxt[6] << endl; // query 3 answered cout << nxt[1] << endl; return 0;} |
2 0 3
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