Count of subarrays starting or ending at an index i such that arr[i] is maximum in subarray
Last Updated :
20 Aug, 2021
Given an array arr[] consisting of N integers, the task is to find the number of subarrays starting or ending at an index i such that arr[i] is the maximum element of the subarray.
Examples:
Input: arr[] = {3, 4, 1, 6, 2}
Output: 1 3 1 5 1
Explanation:
- The subarray starting or ending at index 0 and with maximum arr[0](=3) is {3}. Therefore, the count is 1.
- The subarrays starting or ending at index 1 and with maximum arr[1](=4) are {3, 4}, {4}, and {4, 1}. Therefore, the count is 3.
- The subarray starting or ending at index 2 and with maximum arr[2](=1) is {1}. Therefore, the count is 1.
- The subarrays starting or ending at index 3 and with maximum arr[3](=6) are {3, 4, 1, 6}, {4, 1, 6}, {1, 6}, {6}, and {6, 2}. Therefore, the count is 5.
- The subarray starting or ending at index 4 and with maximum arr[4](=2) is {2}. Therefore, the count is 1.
Input: arr[] = {1, 2, 3}
Output: 1 2 3
Naive Approach: The simplest approach to solve the given problem is for every ith index, iterate backward and forward until the maximum of the subarray remains equal to arr[i] and then print the total count of subarrays obtained.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by storing the index of the next greater element and previous greater element for every index i and find the count of subarrays accordingly for each index. Follow the steps below to solve the given problem:
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find previous greater
// element
int* getPGE(int arr[], int n)
{
// Stores the previous greater
// element for each index
int* pge = new int[n];
stack<int> stack;
// Traverse the array
for(int i = 0; i < n; i++)
{
// Iterate until stack is empty
// and top element is less than
// the current element arr[i]
while (!stack.empty() &&
arr[stack.top()] <= arr[i])
{
stack.pop();
}
// Update the previous greater
// element for arr[i]
pge[i] = stack.empty() ? -1 : stack.top();
// Push the current index to
// the stack
stack.push(i);
}
// Return the PGE[] array
return pge;
}
// Function to find the Next Greater Element
int* getNGE(int arr[], int n)
{
// Stores the next greater element
// for each index
int* nge = new int[n];
stack<int> stack;
// Traverse the array from the back
for(int i = n - 1; i >= 0; i--)
{
// Iterate until stack is empty
// and top element is less than
// the current element arr[i]
while (!stack.empty() &&
arr[stack.top()] <= arr[i])
{
stack.pop();
}
// Update the next greater
// element for arr[i]
nge[i] = stack.empty() ? n : stack.top();
// Push the current index
stack.push(i);
}
// Return the NGE[] array
return nge;
}
// Function to find the count of
// subarrays starting or ending at
// index i having arr[i] as maximum
void countSubarrays(int arr[], int n)
{
// Function call to find the
// previous greater element
// for each array elements
int* pge = getPGE(arr, n);
// Function call to find the
// previous greater element
// for each elements
int* nge = getNGE(arr, n);
// Traverse the array arr[]
for(int i = 0; i < n; i++)
{
// Print count of subarrays
// satisfying the conditions
cout << nge[i] - pge[i] - 1 << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 3, 4, 1, 6, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
countSubarrays(arr, n);
return 0;
}
// This code is contributed by Potta Lokesh
// Java program for the above approach
import java.util.*;
public class GFG {
// Function to find previous greater
// element
private static int[] getPGE(int[] arr)
{
// Stores the previous greater
// element for each index
int[] pge = new int[arr.length];
Stack<Integer> stack = new Stack<>();
// Traverse the array
for (int i = 0; i < arr.length; i++) {
// Iterate until stack is empty
// and top element is less than
// the current element arr[i]
while (!stack.isEmpty()
&& arr[stack.peek()] <= arr[i]) {
stack.pop();
}
// Update the previous greater
// element for arr[i]
pge[i] = stack.isEmpty() ? -1 : stack.peek();
// Push the current index to
// the stacl
stack.push(i);
}
// Return the PGE[] array
return pge;
}
// Function to find the Next Greater Element
private static int[] getNGE(int[] arr)
{
// Stores the next greater element
// for each index
int[] nge = new int[arr.length];
Stack<Integer> stack = new Stack<>();
// Traverse the array from the back
for (int i = arr.length - 1; i >= 0; i--) {
// Iterate until stack is empty
// and top element is less than
// the current element arr[i]
while (!stack.isEmpty()
&& arr[stack.peek()] <= arr[i]) {
stack.pop();
}
// Update the next greater
// element for arr[i]
nge[i] = stack.isEmpty() ? arr.length
: stack.peek();
// Push the current index
stack.push(i);
}
// Return the NGE[] array
return nge;
}
// Function to find the count of
// subarrays starting or ending at
// index i having arr[i] as maximum
private static void countSubarrays(int[] arr)
{
// Function call to find the
// previous greater element
// for each array elements
int[] pge = getPGE(arr);
// Function call to find the
// previous greater element
// for each elements
int[] nge = getNGE(arr);
// Traverse the array arr[]
for (int i = 0; i < arr.length; i++) {
// Print count of subarrays
// satisfying the conditions
System.out.print(
nge[i] - pge[i] - 1 + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = new int[] { 3, 4, 1, 6, 2 };
countSubarrays(arr);
}
}
# Python program for the above approach
# Stores the previous greater
# element for each index
pge = []
# Stores the next greater element
# for each index
nge = []
# Function to find previous greater
# element
def getPGE(arr, n) :
s = list()
# Traverse the array
for i in range(0, n):
# Iterate until stack is empty
# and top element is less than
# the current element arr[i]
while (len(s) > 0 and arr[s[-1]] <= arr[i]):
s.pop()
# Update the previous greater
# element for arr[i]
if len(s) == 0:
pge.append(-1)
else:
pge.append(s[-1])
# Push the current index
s.append(i)
# Function to find the Next Greater Element
def getNGE(arr, n) :
s = list()
# Traverse the array from the back
for i in range(n-1, -1, -1):
# Iterate until stack is empty
# and top element is less than
# the current element arr[i]
while (len(s) > 0 and arr[s[-1]] <= arr[i]):
s.pop()
# Update the next greater
# element for arr[i]
if len(s) == 0:
nge.append(n)
else:
nge.append(s[-1])
# Push the current index
s.append(i)
nge.reverse();
# Function to find the count of
# subarrays starting or ending at
# index i having arr[i] as maximum
def countSubarrays(arr, n):
# Function call to find the
# previous greater element
# for each array elements
getNGE(arr, n);
# Function call to find the
# previous greater element
# for each elements
getPGE(arr, n);
# Traverse the array arr[]
for i in range(0,n):
print(nge[i]-pge[i]-1,end = " ")
arr = [ 3, 4, 1, 6, 2 ]
n = len(arr)
countSubarrays(arr,n);
# This code is contributed by codersaty
// C# program for the above approach
using System;
using System.Collections;
class GFG {
// Function to find previous greater
// element
static int[] getPGE(int[] arr)
{
// Stores the previous greater
// element for each index
int[] pge = new int[arr.Length];
Stack stack = new Stack();
// Traverse the array
for (int i = 0; i < arr.Length; i++) {
// Iterate until stack is empty
// and top element is less than
// the current element arr[i]
while (stack.Count > 0 && arr[(int)stack.Peek()] <= arr[i]) {
stack.Pop();
}
// Update the previous greater
// element for arr[i]
pge[i] = stack.Count == 0 ? -1 : (int)stack.Peek();
// Push the current index to
// the stacl
stack.Push(i);
}
// Return the PGE[] array
return pge;
}
// Function to find the Next Greater Element
static int[] getNGE(int[] arr)
{
// Stores the next greater element
// for each index
int[] nge = new int[arr.Length];
Stack stack = new Stack();
// Traverse the array from the back
for (int i = arr.Length - 1; i >= 0; i--) {
// Iterate until stack is empty
// and top element is less than
// the current element arr[i]
while (stack.Count > 0 && arr[(int)stack.Peek()] <= arr[i]) {
stack.Pop();
}
// Update the next greater
// element for arr[i]
nge[i] = stack.Count == 0 ? arr.Length : (int)stack.Peek();
// Push the current index
stack.Push(i);
}
// Return the NGE[] array
return nge;
}
// Function to find the count of
// subarrays starting or ending at
// index i having arr[i] as maximum
static void countSubarrays(int[] arr)
{
// Function call to find the
// previous greater element
// for each array elements
int[] pge = getPGE(arr);
// Function call to find the
// previous greater element
// for each elements
int[] nge = getNGE(arr);
// Traverse the array arr[]
for (int i = 0; i < arr.Length; i++) {
// Print count of subarrays
// satisfying the conditions
Console.Write((nge[i] - pge[i] - 1) + " ");
}
}
// Driver code
static void Main() {
int[] arr = { 3, 4, 1, 6, 2 };
countSubarrays(arr);
}
}
// This code is contributed by divyesh072019.
<script>
// Javascript program for the above approach
// Stores the previous greater
// element for each index
let pge = [];
// Stores the next greater element
// for each index
let nge = [];
// Function to find previous greater
// element
function getPGE(arr, n) {
let s = [];
// Traverse the array
for (let i = 0; i < n; i++)
{
// Iterate until stack is empty
// and top element is less than
// the current element arr[i]
while (s.length > 0 && arr[s[s.length - 1]] <= arr[i])
{
s.pop();
}
// Update the previous greater
// element for arr[i]
if (s.length == 0) pge.push(-1);
else pge.push(s[s.length - 1]);
// Push the current index
s.push(i);
}
}
// Function to find the Next Greater Element
function getNGE(arr, n) {
let s = [];
// Traverse the array from the back
for (let i = n - 1; i >= 0; i--)
{
// Iterate until stack is empty
// and top element is less than
// the current element arr[i]
while (s.length > 0 && arr[s[s.length - 1]] <= arr[i]) {
s.pop();
}
// Update the next greater
// element for arr[i]
if (s.length == 0) nge.push(n);
else nge.push(s[s.length - 1]);
// Push the current index
s.push(i);
}
nge.reverse();
}
// Function to find the count of
// subarrays starting or ending at
// index i having arr[i] as maximum
function countSubarrays(arr, n)
{
// Function call to find the
// previous greater element
// for each array elements
getNGE(arr, n);
// Function call to find the
// previous greater element
// for each elements
getPGE(arr, n);
// Traverse the array arr[]
for (let i = 0; i < n; i++) {
document.write(nge[i] - pge[i] - 1 + " ");
}
}
let arr = [3, 4, 1, 6, 2];
let n = arr.length;
countSubarrays(arr, n);
// This code is contributed by _saurabh_jaiswal
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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