Longest Subarray with first element greater than or equal to Last
Last Updated :
11 Mar, 2025
Given an array of integers arr[] of size n. Your task is to find the maximum length subarray such that its first element is greater than or equals to the last element.
Examples:
Input : arr[] = [-5, -1, 7, 5, 1, -2]
Output : 5
Explanation : Subarray {-1, 7, 5, 1, -2} forms maximum length subarray with its first element greater than the last.
Input : arr[] = [1, 5, 7]
Output : 1
Explanation: The given array is in strictly increasing order, thus only individuals elements can form such subarrays.
[Naive Approach] - Using Nested Loops - O(n ^ 2) Time and O(1) Space
The idea is to use nested loops, where the outer loop marks the starting of subarray and the inner loop marks the ending, and for each iteration of inner loop, check if the current element is less than or equals to outer loop element, if so update the maximum subarray length accordingly.
Below is given the implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find longest subarray with first
// element greater than or equals to last element
int longestSubarr(vector<int> &arr) {
int n = arr.size();
// to store the answer
int ans = 0;
for(int i = 0; i < n; i++) {
for(int j = i; j < n; j++) {
// check if the first element is greater
// than or equals to the last element
if(arr[i] >= arr[j]) {
ans = max(ans, j - i + 1);
}
}
}
return ans;
}
int main() {
vector<int> arr = { -5, -1, 7, 5, 1, -2 };
cout << longestSubarr(arr);
return 0;
}
// Function to find longest subarray with first
// element greater than or equals to last element
import java.util.*;
class GfG {
// Function to find longest subarray with first
// element greater than or equals to last element
static int longestSubarr(int[] arr) {
int n = arr.length;
// to store the answer
int ans = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
// check if the first element is greater
// than or equals to the last element
if (arr[i] >= arr[j]) {
ans = Math.max(ans, j - i + 1);
}
}
}
return ans;
}
public static void main(String[] args) {
int[] arr = { -5, -1, 7, 5, 1, -2 };
System.out.println(longestSubarr(arr));
}
}
# Function to find longest subarray with first
# element greater than or equals to last element.
def longestSubarr(arr):
n = len(arr)
# to store the answer
ans = 0
for i in range(n):
for j in range(i, n):
# check if the first element is greater
# than or equals to the last element
if arr[i] >= arr[j]:
ans = max(ans, j - i + 1)
return ans
if __name__ == "__main__":
arr = [-5, -1, 7, 5, 1, -2]
print(longestSubarr(arr))
// Function to find longest subarray with first
// element greater than or equals to last element
using System;
using System.Collections.Generic;
class GfG {
// Function to find longest subarray with first
// element greater than or equals to last element
static int longestSubarr(int[] arr) {
int n = arr.Length;
// to store the answer
int ans = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
// check if the first element is greater
// than or equals to the last element
if (arr[i] >= arr[j]) {
ans = Math.Max(ans, j - i + 1);
}
}
}
return ans;
}
static void Main() {
int[] arr = { -5, -1, 7, 5, 1, -2 };
Console.WriteLine(longestSubarr(arr));
}
}
// Function to find longest subarray with first
// element greater than or equals to the last element.
function longestSubarr(arr) {
let n = arr.length;
// to store the answer
let ans = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
// check if the first element is greater
// than or equals to the last element
if (arr[i] >= arr[j])
ans = Math.max(ans, j - i + 1);
}
}
return ans;
}
let arr = [-5, -1, 7, 5, 1, -2];
console.log(longestSubarr(arr));
[Better Approach] - Using Binary Search - O(n * log n) Time and O(n) Space
The idea is to create an auxiliary array ind[] of size n, where ind[i] stores the index of the maximum element in range [0, i], and then perform binary search for each index j of the given array arr[] to find the smallest index i, such that arr[ind[i]] >= arr[j].
Follow the below given steps:
- Create an auxiliary array ind[] of size n, and a counter maxi to store the maximum element.
- Run a loop from 0 to n - 1, and for each index i, if arr[i] > maxi, update maxi and set ind[i] to i, else set ind[i] to ind[i-1].
- Now, again run a loop from 0 to n - 1, and for each element arr[i], find the smallest index j, such that arr[ind[j]] >= arr[i] using binary search.
- Update the answer to the max of answer and i - ind[j] + 1.
Below is given the implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the lowest index j such that
// arr[j] >= arr[i] using binary search
int findIndex(vector<int> &arr, vector<int> &ind, int i) {
int low = 0, high = i;
while(low < high) {
int mid = low + (high - low) / 2;
// if arr[ind[mid]] >= arr[i]
// then search in the left half
if(arr[ind[mid]] >= arr[i]) {
high = mid;
}
// else search in the right half
else {
low = mid + 1;
}
}
return ind[low];
}
// Function to find longest subarray with first
// element greater than or equals to last element
int longestSubarr(vector<int> &arr) {
int n = arr.size();
// create the array ind[]
vector<int> ind(n);
// to store the maximum element
int maxi = INT_MIN;
for(int i = 0; i < n; i++) {
// maxi is smaller than arr[i]
// then update maxi and ind[i]
if(maxi < arr[i]) {
maxi = arr[i];
ind[i] = i;
}
// else set ind[i] equal to previous
else {
ind[i] = ind[i - 1];
}
}
// to store the answer
int ans = 0;
for(int i = 0; i < n; i++) {
// find the lowest index j such that
// arr[j] >= arr[i] using binary search
int index = findIndex(arr, ind, i);
// update the answer
ans = max(ans, i - index + 1);
}
return ans;
}
int main() {
vector<int> arr = { -5, -1, 7, 5, 1, -2 };
cout << longestSubarr(arr);
return 0;
}
// Function to find the lowest index j such that
// arr[j] >= arr[i] using binary search
import java.util.*;
class GfG {
// Function to find the lowest index j such that
// arr[j] >= arr[i] using binary search
static int findIndex(int[] arr, int[] ind, int i) {
int low = 0, high = i;
while (low < high) {
int mid = low + (high - low) / 2;
// if arr[ind[mid]] >= arr[i]
// then search in the left half
if (arr[ind[mid]] >= arr[i]) {
high = mid;
}
// else search in the right half
else {
low = mid + 1;
}
}
return ind[low];
}
// Function to find longest subarray with first
// element greater than or equals to last element
static int longestSubarr(int[] arr) {
int n = arr.length;
// create the array ind[]
int[] ind = new int[n];
// to store the maximum element
int maxi = Integer.MIN_VALUE;
for (int i = 0; i < n; i++) {
// maxi is smaller than arr[i]
// then update maxi and ind[i]
if (maxi < arr[i]) {
maxi = arr[i];
ind[i] = i;
}
// else set ind[i] equal to previous
else {
ind[i] = ind[i - 1];
}
}
// to store the answer
int ans = 0;
for (int i = 0; i < n; i++) {
// find the lowest index j such that
// arr[j] >= arr[i] using binary search
int index = findIndex(arr, ind, i);
// update the answer
ans = Math.max(ans, i - index + 1);
}
return ans;
}
public static void main(String[] args) {
int[] arr = { -5, -1, 7, 5, 1, -2 };
System.out.println(longestSubarr(arr));
}
}
# Function to find the lowest index j such that
# arr[j] >= arr[i] using binary search.
def findIndex(arr, ind, i):
low = 0
high = i
while low < high:
mid = low + (high - low) // 2
# if arr[ind[mid]] >= arr[i]
# then search in the left half
if arr[ind[mid]] >= arr[i]:
high = mid
else:
low = low + 1 if (low + 1) == high else mid + 1
return ind[low]
# Function to find longest subarray with first
# element greater than or equals to last element.
def longestSubarr(arr):
n = len(arr)
# create the array ind[]
ind = [0] * n
# to store the maximum element
maxi = -float('inf')
for i in range(n):
# maxi is smaller than arr[i]
# then update maxi and ind[i]
if maxi < arr[i]:
maxi = arr[i]
ind[i] = i
# else set ind[i] equal to previous
else:
ind[i] = ind[i - 1]
# to store the answer
ans = 0
for i in range(n):
# find the lowest index j such that
# arr[j] >= arr[i] using binary search
index = findIndex(arr, ind, i)
# update the answer
ans = max(ans, i - index + 1)
return ans
if __name__ == "__main__":
arr = [-5, -1, 7, 5, 1, -2]
print(longestSubarr(arr))
// Function to find the lowest index j such that
// arr[j] >= arr[i] using binary search
using System;
using System.Collections.Generic;
class GfG {
// Function to find the lowest index j such that
// arr[j] >= arr[i] using binary search
static int findIndex(int[] arr, int[] ind, int i) {
int low = 0, high = i;
while (low < high) {
int mid = low + (high - low) / 2;
// if arr[ind[mid]] >= arr[i]
// then search in the left half
if (arr[ind[mid]] >= arr[i])
high = mid;
else
low = mid + 1;
}
return ind[low];
}
// Function to find longest subarray with first
// element greater than or equals to last element
static int longestSubarr(int[] arr) {
int n = arr.Length;
// create the array ind[]
int[] ind = new int[n];
// to store the maximum element
int maxi = int.MinValue;
for (int i = 0; i < n; i++) {
// maxi is smaller than arr[i]
// then update maxi and ind[i]
if (maxi < arr[i]) {
maxi = arr[i];
ind[i] = i;
}
// else set ind[i] equal to previous
else {
ind[i] = ind[i - 1];
}
}
// to store the answer
int ans = 0;
for (int i = 0; i < n; i++) {
// find the lowest index j such that
// arr[j] >= arr[i] using binary search
int index = findIndex(arr, ind, i);
// update the answer
ans = Math.Max(ans, i - index + 1);
}
return ans;
}
static void Main() {
int[] arr = { -5, -1, 7, 5, 1, -2 };
Console.WriteLine(longestSubarr(arr));
}
}
// Function to find the lowest index j such that
// arr[j] >= arr[i] using binary search.
function findIndex(arr, ind, i) {
let low = 0, high = i;
while (low < high) {
let mid = low + Math.floor((high - low) / 2);
// if arr[ind[mid]] >= arr[i]
// then search in the left half
if (arr[ind[mid]] >= arr[i]) {
high = mid;
} else {
low = mid + 1;
}
}
return ind[low];
}
// Function to find longest subarray with first
// element greater than or equals to last element.
function longestSubarr(arr) {
let n = arr.length;
// create the array ind[]
let ind = new Array(n).fill(0);
// to store the maximum element
let maxi = -Infinity;
for (let i = 0; i < n; i++) {
// maxi is smaller than arr[i]
// then update maxi and ind[i]
if (maxi < arr[i]) {
maxi = arr[i];
ind[i] = i;
} else {
// else set ind[i] equal to previous
ind[i] = ind[i - 1];
}
}
// to store the answer
let ans = 0;
for (let i = 0; i < n; i++) {
// find the lowest index j such that
// arr[j] >= arr[i] using binary search
let index = findIndex(arr, ind, i);
// update the answer
ans = Math.max(ans, i - index + 1);
}
return ans;
}
let arr = [-5, -1, 7, 5, 1, -2];
console.log(longestSubarr(arr));
[Expected Approach] - Using Stack - O(n) Time and O(n) Space
This problem is mainly a variation of previous greater element.
In the above approach, we are using an auxiliary array to store the indices of required elements, but instead of doing so we can use a stack to store indices i such that arr[i] is maximum in all elements from arr[0] to arr[i]. And thereafter, run a loop from reverse and for each index i, pop the element from stack while arr[i] <= arr[st.top()], and update the answer accordingly.
Follow the below given steps:
- Create a stack st to store the required indices.
- Run a loop from 0 to n - 1, and for each index i, if stack is empty of arr[st.top()] < arr[i], store i in the stack st.
- Now run a loop from n - 1 to 0, and for each index i, pop element from stack st while arr[st.top()] >= arr[i].
- Update the answer to the max of answer and i - st.top().
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find longest subarray with first
// element greater than or equals to last element
int longestSubarr(vector<int> &arr) {
int n = arr.size();
// create stack to store the indices
stack<int> st;
// to store the maximum element
int maxi = INT_MIN;
for(int i = 0; i < n; i++) {
// maxi is smaller than arr[i]
// then update maxi and push i into stack
if(maxi < arr[i]) {
maxi = arr[i];
st.push(i);
}
}
// to store the answer
int ans = 0;
for(int i = n - 1; i >= 0; i--) {
// check if the first element is greater
// than or equals to the last element
while(!st.empty() && arr[st.top()] >= arr[i]) {
st.pop();
}
// if stack is empty
if(st.empty()) {
ans = max(ans, i + 1);
}
else {
ans = max(ans, i - st.top());
}
}
return ans;
}
int main() {
vector<int> arr = { -5, -1, 7, 5, 1, -2 };
cout << longestSubarr(arr);
return 0;
}
// Function to find longest subarray with first
// element greater than or equals to last element
import java.util.*;
class GfG {
// Function to find longest subarray with first
// element greater than or equals to last element
static int longestSubarr(int[] arr) {
int n = arr.length;
// create stack to store the indices
Stack<Integer> st = new Stack<>();
// to store the maximum element
int maxi = Integer.MIN_VALUE;
for (int i = 0; i < n; i++) {
// maxi is smaller than arr[i]
// then update maxi and push i into stack
if (maxi < arr[i]) {
maxi = arr[i];
st.push(i);
}
}
// to store the answer
int ans = 0;
for (int i = n - 1; i >= 0; i--) {
// check if the first element is greater
// than or equals to the last element
while (!st.empty() && arr[st.peek()] >= arr[i]) {
st.pop();
}
// if stack is empty
if (st.empty()) {
ans = Math.max(ans, i + 1);
}
else {
ans = Math.max(ans, i - st.peek());
}
}
return ans;
}
public static void main(String[] args) {
int[] arr = { -5, -1, 7, 5, 1, -2 };
System.out.println(longestSubarr(arr));
}
}
# Function to find longest subarray with first
# element greater than or equals to last element.
def longestSubarr(arr):
n = len(arr)
# create stack to store the indices
st = []
# to store the maximum element
maxi = -float('inf')
for i in range(n):
# maxi is smaller than arr[i]
# then update maxi and push i into stack
if maxi < arr[i]:
maxi = arr[i]
st.append(i)
# to store the answer
ans = 0
for i in range(n - 1, -1, -1):
# check if the first element is greater
# than or equals to the last element
while st and arr[st[-1]] >= arr[i]:
st.pop()
# if stack is empty
if not st:
ans = max(ans, i + 1)
else:
ans = max(ans, i - st[-1])
return ans
if __name__ == "__main__":
arr = [ -5, -1, 7, 5, 1, -2 ]
print(longestSubarr(arr))
// Function to find longest subarray with first
// element greater than or equals to last element
using System;
using System.Collections.Generic;
class GfG {
// Function to find longest subarray with first
// element greater than or equals to last element
static int longestSubarr(int[] arr) {
int n = arr.Length;
// create stack to store the indices
Stack<int> st = new Stack<int>();
// to store the maximum element
int maxi = int.MinValue;
for (int i = 0; i < n; i++) {
// maxi is smaller than arr[i]
// then update maxi and push i into stack
if (maxi < arr[i]) {
maxi = arr[i];
st.Push(i);
}
}
// to store the answer
int ans = 0;
for (int i = n - 1; i >= 0; i--) {
// check if the first element is greater
// than or equals to the last element
while (st.Count > 0 && arr[st.Peek()] >= arr[i]) {
st.Pop();
}
// if stack is empty
if (st.Count == 0) {
ans = Math.Max(ans, i + 1);
}
else {
ans = Math.Max(ans, i - st.Peek());
}
}
return ans;
}
static void Main() {
int[] arr = { -5, -1, 7, 5, 1, -2 };
Console.WriteLine(longestSubarr(arr));
}
}
// Function to find longest subarray with first
// element greater than or equals to last element.
function longestSubarr(arr) {
let n = arr.length;
// create stack to store the indices
let st = [];
// to store the maximum element
let maxi = -Infinity;
for (let i = 0; i < n; i++) {
// maxi is smaller than arr[i]
// then update maxi and push i into stack
if (maxi < arr[i]) {
maxi = arr[i];
st.push(i);
}
}
// to store the answer
let ans = 0;
for (let i = n - 1; i >= 0; i--) {
// check if the first element is greater
// than or equals to the last element
while (st.length > 0 && arr[st[st.length - 1]] >= arr[i]) {
st.pop();
}
// if stack is empty
if (st.length === 0) {
ans = Math.max(ans, i + 1);
} else {
ans = Math.max(ans, i - st[st.length - 1]);
}
}
return ans;
}
let arr = [ -5, -1, 7, 5, 1, -2 ];
console.log(longestSubarr(arr));
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