Given an array of n elements and q queries, for each query which has an 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 then 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 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 ext 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 index 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.
br>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 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.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[] 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 |
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 |
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)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
leave a comment
0 Comments