Largest interval in an Array that contains the given element X for Q queries
Last Updated :
20 May, 2021
Given an array arr[] of N elements and Q queries of the form [X]. For each query, the task is to find the largest interval [L, R] of the array such that the greatest element in the interval is arr[X], such that 1 ? L ? X ? R.
Note: The array has 1-based indexing.
Examples:
Input: N = 5, arr[] = {2, 1, 2, 3, 2}, Q = 3, query[] = {1, 2, 4}
Output:
[1, 3]
[2, 2]
[1, 5]
Explanation :
In 1st query, x = 1, so arr[x] = 2 and answer is L = 1 and R = 3. here, we can see that max(arr[1], arr[2], arr[3]) = arr[x], which is the maximum intervals.
In 2nd query, x = 2, so arr[x] = 1 and since it is the smallest element of the array, so the interval contains only one element, thus the range is [2, 2].
In 3rd query, x = 4, so arr[x] = 4, which is maximum element of the arr[], so the answer is whole array, L = 1 and R = N.
Input: N = 4, arr[] = { 1, 2, 2, 4}, Q = 2, query[] = {1, 2}
Output:
[1, 1]
[1, 3]
Explanation:
In 1st query, x = 1, so arr[x] = 1 and since it is the smallest element of the array, so the interval contains only one element, thus the range is [1, 1].
In 2nd query, x = 2, so arr[x] = 2 and answer is L = 1 and R = 3. here, we can see that max(arr[1], arr[2], arr[3]) = arr[x] = arr[2] = 2, which is the maximum intervals.
Approach: The idea is to precompute the largest interval for every value K in arr[] from 1 to N. Below are the steps:
- For each element K in arr[], fix the index of the element K, then find how much we can extend the interval to it's left and right.
- Decrement left iterator till arr[left] ? K and similarly increment right iterator till arr[right] ? K.
- The final value of left and right represents the starting and the ending index of the interval, which is stored in arrL[] and arrR[] respectively.
- After we have precomputed interval range for each value. Then, for each query, we need to print the interval range for arr[x] i.e., arrL[arr[x]] and arrR[arr[x]].
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to precompute the interval
// for each query
void utilLargestInterval(int arr[],
int arrL[],
int arrR[],
int N)
{
// For every values [1, N] find
// the longest intervals
for (int maxValue = 1;
maxValue <= N; maxValue++) {
int lastIndex = 0;
// Iterate the array arr[]
for (int i = 1; i <= N; i++) {
if (lastIndex >= i
|| arr[i] != maxValue)
continue;
int left = i, right = i;
// Shift the left pointers
while (left > 0
&& arr[left] <= maxValue)
left--;
// Shift the right pointers
while (right <= N
&& arr[right] <= maxValue)
right++;
left++, right--;
lastIndex = right;
// Store the range of interval
// in arrL[] and arrR[].
for (int j = left; j <= right; j++) {
if (arr[j] == maxValue) {
arrL[j] = left;
arrR[j] = right;
}
}
}
}
}
// Function to find the largest interval
// for each query in Q[]
void largestInterval(
int arr[], int query[], int N, int Q)
{
// To store the L and R of X
int arrL[N + 1], arrR[N + 1];
// Function Call
utilLargestInterval(arr, arrL,
arrR, N);
// Iterate to find ranges for each query
for (int i = 0; i < Q; i++) {
cout << "[" << arrL[query[i]]
<< ", " << arrR[query[i]]
<< "]\n";
}
}
// Driver Code
int main()
{
int N = 5, Q = 3;
// Given array arr[]
int arr[N + 1] = { 0, 2, 1, 2, 3, 2 };
// Given Queries
int query[Q] = { 1, 2, 4 };
// Function Call
largestInterval(arr, query, N, Q);
return 0;
}
// Java program for the above approach
class GFG{
// Function to precompute the interval
// for each query
static void utilLargestInterval(int arr[],
int arrL[],
int arrR[],
int N)
{
// For every values [1, N] find
// the longest intervals
for(int maxValue = 1;
maxValue <= N; maxValue++)
{
int lastIndex = 0;
// Iterate the array arr[]
for(int i = 1; i <= N; i++)
{
if (lastIndex >= i ||
arr[i] != maxValue)
continue;
int left = i, right = i;
// Shift the left pointers
while (left > 0 &&
arr[left] <= maxValue)
left--;
// Shift the right pointers
while (right <= N &&
arr[right] <= maxValue)
right++;
left++;
right--;
lastIndex = right;
// Store the range of interval
// in arrL[] and arrR[].
for(int j = left; j <= right; j++)
{
if (arr[j] == maxValue)
{
arrL[j] = left;
arrR[j] = right;
}
}
}
}
}
// Function to find the largest interval
// for each query in Q[]
static void largestInterval(int arr[],
int query[],
int N, int Q)
{
// To store the L and R of X
int []arrL = new int[N + 1];
int []arrR = new int[N + 1];
// Function Call
utilLargestInterval(arr, arrL,
arrR, N);
// Iterate to find ranges for
// each query
for(int i = 0; i < Q; i++)
{
System.out.print("[" + arrL[query[i]] +
", " + arrR[query[i]] + "]\n");
}
}
// Driver Code
public static void main(String[] args)
{
int N = 5, Q = 3;
// Given array arr[]
int arr[] = { 0, 2, 1, 2, 3, 2 };
// Given queries
int query[] = { 1, 2, 4 };
// Function call
largestInterval(arr, query, N, Q);
}
}
// This code is contributed by Amit Katiyar
# Python3 program for the above approach
# Function to precompute the interval
# for each query
def utilLargestInterval(arr, arrL, arrR, N):
# For every values [1, N] find
# the longest intervals
for maxValue in range(1, N + 1):
lastIndex = 0
# Iterate the array arr[]
for i in range(N + 1):
if (lastIndex >= i or
arr[i] != maxValue):
continue
left = i
right = i
# Shift the left pointers
while (left > 0 and
arr[left] <= maxValue):
left -= 1
# Shift the right pointers
while (right <= N and
arr[right] <= maxValue):
right += 1
left += 1
right -= 1
lastIndex = right
# Store the range of interval
# in arrL[] and arrR[].
for j in range(left, right + 1):
if (arr[j] == maxValue):
arrL[j] = left
arrR[j] = right
# Function to find the largest interval
# for each query in Q[]
def largestInterval(arr, query, N, Q):
# To store the L and R of X
arrL = [0 for i in range(N + 1)]
arrR = [0 for i in range(N + 1)]
# Function call
utilLargestInterval(arr, arrL, arrR, N);
# Iterate to find ranges for each query
for i in range(Q):
print('[' + str(arrL[query[i]]) +
', ' + str(arrR[query[i]]) + ']')
# Driver code
if __name__=="__main__":
N = 5
Q = 3
# Given array arr[]
arr = [ 0, 2, 1, 2, 3, 2 ]
# Given Queries
query = [ 1, 2, 4 ]
# Function call
largestInterval(arr, query, N, Q)
# This code is contributed by rutvik_56
// C# program for the above approach
using System;
class GFG{
// Function to precompute the interval
// for each query
static void utilLargestInterval(int []arr,
int []arrL,
int []arrR,
int N)
{
// For every values [1, N] find
// the longest intervals
for(int maxValue = 1;
maxValue <= N; maxValue++)
{
int lastIndex = 0;
// Iterate the array []arr
for(int i = 1; i <= N; i++)
{
if (lastIndex >= i ||
arr[i] != maxValue)
continue;
int left = i, right = i;
// Shift the left pointers
while (left > 0 &&
arr[left] <= maxValue)
left--;
// Shift the right pointers
while (right <= N &&
arr[right] <= maxValue)
right++;
left++;
right--;
lastIndex = right;
// Store the range of interval
// in arrL[] and arrR[].
for(int j = left; j <= right; j++)
{
if (arr[j] == maxValue)
{
arrL[j] = left;
arrR[j] = right;
}
}
}
}
}
// Function to find the largest interval
// for each query in Q[]
static void largestInterval(int []arr,
int []query,
int N, int Q)
{
// To store the L and R of X
int []arrL = new int[N + 1];
int []arrR = new int[N + 1];
// Function Call
utilLargestInterval(arr, arrL,
arrR, N);
// Iterate to find ranges for
// each query
for(int i = 0; i < Q; i++)
{
Console.Write("[" + arrL[query[i]] +
", " + arrR[query[i]] + "]\n");
}
}
// Driver Code
public static void Main(String[] args)
{
int N = 5, Q = 3;
// Given array []arr
int []arr = { 0, 2, 1, 2, 3, 2 };
// Given queries
int []query = { 1, 2, 4 };
// Function call
largestInterval(arr, query, N, Q);
}
}
// This code is contributed by Princi Singh
<script>
// Javascript program for the above approach
// Function to precompute the interval
// for each query
function utilLargestInterval(arr, arrL, arrR, N)
{
// For every values [1, N] find
// the longest intervals
for (var maxValue = 1;
maxValue <= N; maxValue++) {
var lastIndex = 0;
// Iterate the array arr[]
for (var i = 1; i <= N; i++) {
if (lastIndex >= i
|| arr[i] != maxValue)
continue;
var left = i, right = i;
// Shift the left pointers
while (left > 0
&& arr[left] <= maxValue)
left--;
// Shift the right pointers
while (right <= N
&& arr[right] <= maxValue)
right++;
left++, right--;
lastIndex = right;
// Store the range of interval
// in arrL[] and arrR[].
for (var j = left; j <= right; j++) {
if (arr[j] == maxValue) {
arrL[j] = left;
arrR[j] = right;
}
}
}
}
}
// Function to find the largest interval
// for each query in Q[]
function largestInterval( arr, query, N, Q)
{
// To store the L and R of X
var arrL = Array(N+1).fill(0),arrR = Array(N+1).fill(0);
// Function Call
utilLargestInterval(arr, arrL,
arrR, N);
// Iterate to find ranges for each query
for (var i = 0; i < Q; i++) {
document.write( "[" + arrL[query[i]]
+ ", " + arrR[query[i]]
+ "]<br>");
}
}
// Driver Code
var N = 5, Q = 3;
// Given array arr[]
var arr = [0, 2, 1, 2, 3, 2];
// Given Queries
var query = [1, 2, 4];
// Function Call
largestInterval(arr, query, N, Q);
// This code is contributed by itsok.
</script>
Output: [1, 3]
[2, 2]
[1, 5]
Time Complexity: O(Q + N2)
Auxiliary Space: O(N)
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