Number of NGEs to the right

Software Development
#1

Hello Everyone,

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 won’t 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 won’t 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.

#include <bits/stdc++.h>

using namespace std;

// array to store the next greater element index

void 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 right

void 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 function

int 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;

}

Output

2 0 3

Time complexity: O(1) to answer a query.
Auxiliary Space: O(n)