Longest Subarray with first element greater than or equal to Last
Last Updated :
11 Jul, 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));
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem