First subarray with negative sum from the given Array
Last Updated :
04 Dec, 2023
Given an array arr[] consisting of N integers, the task is to find the start and end indices of the first subarray with a Negative Sum. Print "-1" if no such subarray exists. Note: In the case of multiple negative-sum subarrays in the given array, the first subarray refers to the subarray with the lowest starting index.
Examples:
Input: arr[] = {3, 3, -4, -2} Output: 1 2 Explanation: The first subarray with negative sum is from index 1 to 2 that is arr[1] + arr[2] = -1. Input: arr[] = {1, 2, 3, 10}. Output: -1 Explanation: No Subarray with negative sum exists.
Naive Approach: The naive approach is to generate all subarrays from left to right in the array and check whether any of these subarrays have a negative sum or not. If yes then print the starting and ending index of that subarray.
Steps to implement-
- Declare a vector "ans" to store the answer
- Run two loops to find all subarrays
- Simultaneously find the sum of all elements of the subarray
- If the sum of all elements of the subarray became negative then push starting and last index of the subarray into the vector and return/print that
- In the last, if nothing got printed or returned then return or print "-1"
Code-
C++
// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the index
// of first negative subarray sum
vector<int> findSubarray(int arr[], int N)
{
//To store answer
vector<int> ans;
//Find all subarray
for(int i=0;i<N;i++){
//To store sum of subarray
int sum=0;
for(int j=i;j<N;j++){
//Take this element in finding sum of subarray
sum+=arr[j];
//If subarray has negative sum then store
//its starting and last index in the ans vector
if(sum<0){
ans.push_back(i);
ans.push_back(j);
return ans;
}
}
}
//If any subarray sum is not negative
ans.push_back(-1);
return ans;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 2, -1, 3, -4, 3, -5 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
vector<int> res = findSubarray(arr, n);
// If subarray does not exist
if (res[0] == -1)
cout << "-1" << endl;
// If the subarray exists
else {
cout << res[0]
<< " " << res[1];
}
return 0;
}
import java.util.ArrayList;
import java.util.List;
public class Main {
// Function to return the index of the first negative subarray sum
static List<Integer> findSubarray(int[] arr, int N) {
// To store the answer
List<Integer> ans = new ArrayList<>();
// Find all subarrays
for (int i = 0; i < N; i++) {
// To store the sum of subarray
int sum = 0;
for (int j = i; j < N; j++) {
// Take this element in finding the sum of subarray
sum += arr[j];
// If the subarray has a negative sum, then store
// its starting and last index in the ans list
if (sum < 0) {
ans.add(i);
ans.add(j);
return ans;
}
}
}
// If no subarray sum is negative
ans.add(-1);
return ans;
}
// Driver Code
public static void main(String[] args) {
// Given array arr[]
int arr[] = { 1, 2, -1, 3, -4, 3, -5 };
int n = arr.length;
// Function Call
List<Integer> res = findSubarray(arr, n);
// If subarray does not exist
if (res.get(0) == -1)
System.out.println("-1");
// If the subarray exists
else {
System.out.println(res.get(0) + " " + res.get(1));
}
}
}
def find_subarray(arr):
# To store answer
ans = []
N = len(arr)
# Find all subarrays
for i in range(N):
# To store sum of subarray
subarray_sum = 0
for j in range(i, N):
# Take this element in finding sum of subarray
subarray_sum += arr[j]
# If subarray has negative sum, then store its starting and last index in the ans list
if subarray_sum < 0:
ans.append(i)
ans.append(j)
return ans
# If any subarray sum is not negative
ans.append(-1)
return ans
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [1, 2, -1, 3, -4, 3, -5]
# Function Call
res = find_subarray(arr)
# If subarray does not exist
if res[0] == -1:
print("-1")
# If the subarray exists
else:
print(res[0], res[1])
using System;
using System.Collections.Generic;
class Program
{
// Function to return the index
// of the first negative subarray sum
static List<int> FindSubarray(int[] arr, int N)
{
// To store the answer
List<int> ans = new List<int>();
// Find all subarrays
for (int i = 0; i < N; i++)
{
// To store the sum of the subarray
int sum = 0;
for (int j = i; j < N; j++)
{
// Take this element in finding the sum of the subarray
sum += arr[j];
// If the subarray has a negative sum, then store
// its starting and last index in the ans list
if (sum < 0)
{
ans.Add(i);
ans.Add(j);
return ans;
}
}
}
// If any subarray sum is not negative
ans.Add(-1);
return ans;
}
// Driver Code
static void Main()
{
// Given array arr[]
int[] arr = { 1, 2, -1, 3, -4, 3, -5 };
int n = arr.Length;
// Function Call
List<int> res = FindSubarray(arr, n);
// If the subarray does not exist
if (res[0] == -1)
Console.WriteLine("-1");
// If the subarray exists
else
{
Console.WriteLine(res[0] + " " + res[1]);
}
}
}
function findSubarray(arr) {
let ans = [];
for (let i = 0; i < arr.length; i++) {
let sum = 0;
for (let j = i; j < arr.length; j++) {
sum += arr[j];
// If subarray has negative sum
if (sum < 0) {
ans.push(i);
ans.push(j);
return ans;
}
}
}
// If any subarray sum is not negative
ans.push(-1);
return ans;
}
// Given array arr
let arr = [1, 2, -1, 3, -4, 3, -5];
// Function call
let res = findSubarray(arr);
// If subarray does not exist
if (res[0] === -1)
console.log("-1");
// If the subarray exists
else {
console.log(res[0] + " " + res[1]);
}
Output-
0 6
Time Complexity: O(N2), because of two nested loops to find all subarray
Auxiliary Space: O(1), because no extra space has been used
Efficient Approach: The idea is to solve the problem using Prefix Sum and Hashing. Below are the steps:
- Calculate the Prefix Sum of the array and store it in HashMap.
- Iterate through the array and for every ith index, where i ranges from [0, N - 1], check if the element at the ith index is negative or not. If so, then arr[i] is the required subarray.
- Otherwise, find an index starting from i + 1, where the prefix sum is smaller than the prefix sum up to i.
- If any such index is found in the above step, then the subarray from indices {i, index} gives the first negative subarray.
- If no such subarray is found, print "-1". Otherwise, print the obtained subarray.
Below is the implementation of the above approach:
C++
// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if a sum less
// than pre_sum is present
int b_search(int pre_sum,
map<int, vector<int> >& m,
int index)
{
// Returns an iterator either equal
// to given key or next greater
auto it = m.lower_bound(pre_sum);
if (it == m.begin())
return -1;
// Decrement the iterator
it--;
// Check if the sum found at
// a greater index
auto it1
= lower_bound(it->second.begin(),
it->second.end(),
index);
if (*it1 > index)
return *it1;
return -1;
}
// Function to return the index
// of first negative subarray sum
vector<int> findSubarray(int arr[], int n)
{
// Stores the prefix sum- index
// mappings
map<int, vector<int> > m;
int sum = 0;
// Stores the prefix sum of
// the original array
int prefix_sum[n];
for (int i = 0; i < n; i++) {
sum += arr[i];
// Check if we get sum negative
// starting from first element
if (sum < 0)
return { 0, i };
prefix_sum[i] = sum;
m[sum].push_back(i);
}
// Iterate through array find
// the sum which is just less
// then the previous prefix sum
for (int i = 1; i < n; i++) {
// Check if the starting element
// is itself negative
if (arr[i] < 0)
// arr[i] becomes the required
// subarray
return { i, i };
else {
int pre_sum = prefix_sum[i - 1];
// Find the index which forms
// negative sum subarray
// from i-th index
int index = b_search(pre_sum,
m, i);
// If a subarray is found
// starting from i-th index
if (index != -1)
return { i, index };
}
}
// Return -1 if no such
// subarray is present
return { -1 };
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 2, -1, 3, -4, 3, -5 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
vector<int> res = findSubarray(arr, n);
// If subarray does not exist
if (res[0] == -1)
cout << "-1" << endl;
// If the subarray exists
else {
cout << res[0]
<< " " << res[1];
}
return 0;
}
/*package whatever //do not write package name here */
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// lower bound for a map's key
public static int
lowerBound(Map<Integer, List<Integer> > m, int pre_sum)
{
int ans = -1;
for (Integer key : m.keySet()) {
if (key >= pre_sum) {
ans = key;
break;
}
}
return ans;
}
// lower bound for a list
public static int lowerBoundList(List<Integer> li,
int target)
{
int ans = -1;
for (int i = 0; i < li.size(); i++) {
if (li.get(i) >= target) {
ans = i;
break;
}
}
return ans;
}
// Function to check if a sum less
// than pre_sum is present
public static int
b_search(int pre_sum, Map<Integer, List<Integer> > m,
int index)
{
// Returns an iterator either equal
// to given key or next greater
int it = lowerBound(m, pre_sum);
if (it == 0)
return -1;
// Decrement the iterator
it--;
List<Integer> map_list = new ArrayList<Integer>();
map_list = m.get(it);
// Check if the sum found at
// a greater index
int it1 = lowerBoundList(map_list, index);
if (map_list.get(it1) > index)
return map_list.get(it1);
return -1;
}
// Function to return the index
// of first negative subarray sum
public static List<Integer> findSubarray(int[] arr,
int n)
{
// Stores the prefix sum- index
// mappings
Map<Integer, List<Integer> > m
= new HashMap<Integer, List<Integer> >();
for (int i = 0; i < n; i++) {
List<Integer> a = new ArrayList<Integer>();
a.add(0);
m.put(i, a);
}
int sum = 0;
// Stores the prefix sum of
// the original array
int[] prefix_sum = new int[n];
for (int i = 0; i < n; i++) {
sum += arr[i];
// Check if we get sum negative
// starting from first element
if (sum < 0) {
List<Integer> xyz
= new ArrayList<Integer>();
xyz.add(0);
xyz.add(i);
return xyz;
// return { 0, i };
}
List<Integer> xyz = new ArrayList<Integer>();
xyz.add(i);
prefix_sum[i] = sum;
m.put(sum, xyz);
}
// Iterate through array find
// the sum which is just less
// then the previous prefix sum
for (int i = 1; i < n; i++)
{
// Check if the starting element
// is itself negative
if (arr[i] < 0)
{
// arr[i] becomes the required
// subarray
List<Integer> ret
= new ArrayList<Integer>();
ret.add(i);
ret.add(i);
return ret;
// return { i, i };
}
else {
int pre_sum = prefix_sum[i - 1];
// Find the index which forms
// negative sum subarray
// from i-th index
int index = b_search(pre_sum, m, i);
// If a subarray is found
// starting from i-th index
if (index != -1) {
List<Integer> ret
= new ArrayList<Integer>();
ret.add(i);
ret.add(index);
return ret;
// return { i, index };
}
}
}
// Return -1 if no such
// subarray is present
List<Integer> re = new ArrayList<Integer>();
re.add(-1);
return re;
}
public static void main(String[] args)
{
// Given array arr[]
int[] arr = { 1, 2, -1, 3, -4, 3, -5 };
int n = arr.length;
// Function Call
List<Integer> res = new ArrayList<Integer>();
res = findSubarray(arr, n);
// If subarray does not exist
if (res.get(0) == -1)
System.out.println("-1");
// If the subarray exists
else {
System.out.print(res.get(0));
System.out.print(" ");
System.out.println(res.get(1));
}
}
}
// This code is contributed by akashish__
# lower bound for a dictionary's key
def lowerBound(m, pre_sum):
ans = -1
for key in m:
if (key >= pre_sum):
ans = key
break
return ans
# lower bound for a list
def lowerBoundList(li, target):
ans = -1
for i in range(0,len(li)):
if (li[i] >= target):
ans = i
break
return ans
# Function to check if a sum less
# than pre_sum is present
def b_search(pre_sum, m, index):
# Returns an iterator either equal
# to given key or next greater
it = lowerBound(m, pre_sum)
if (it == 0):
return -1
# Decrement the iterator
it = it - 1
map_list = m[it]
# Check if the sum found at
# a greater index
it1 = lowerBoundList(map_list, index)
if (map_list[it1] > index):
return map_list[it1]
return -1
# Function to return the index
# of first negative subarray sum
def findSubarray(arr, n):
# Stores the prefix sum- index
# mappings
m = {}
for i in range(0,n):
a = [0]
m[i] = a
sum = 0
# Stores the prefix sum of
# the original array
prefix_sum = [0]*n
for i in range(0,n):
sum += arr[i]
# Check if we get sum negative
# starting from first element
if (sum < 0):
xyz = [0,i]
return xyz
xyz = [i]
prefix_sum[i] = sum
m[sum] = xyz
# Iterate through array find
# the sum which is just less
# then the previous prefix sum
for i in range(1,n):
# Check if the starting element
# is itself negative
if (arr[i] < 0):
# arr[i] becomes the required
# subarray
ret = [i,i]
return ret
else:
pre_sum = prefix_sum[i - 1]
# Find the index which forms
# negative sum subarray
# from i-th index
index = b_search(pre_sum, m, i)
# If a subarray is found
# starting from i-th index
if (index != -1):
ret = [i,index]
return ret
# Return -1 if no such
# subarray is present
re = [-1]
return re
# Given array arr[]
arr = [ 1, 2, -1, 3, -4, 3, -5 ]
n = len(arr)
# Function Call
res = findSubarray(arr, n)
# If subarray does not exist
if (res[0] == -1):
print("-1")
# If the subarray exists
else:
print(res)
# This code is contributed by akashish__
using System;
using System.Collections.Generic;
public class GFG
{
// lower bound for a dictionary's key
public static int
lowerBound(Dictionary<int, List<int> > m, int pre_sum)
{
int ans = -1;
foreach(KeyValuePair<int, List<int> > kvp in m)
{
if (kvp.Key >= pre_sum) {
ans = kvp.Key;
break;
}
}
return ans;
}
// lower bound for a list
public static int lowerBoundList(List<int> li,
int target)
{
int ans = -1;
for (int i = 0; i < li.Count; i++) {
if (li[i] >= target) {
ans = i;
break;
}
}
return ans;
}
// Function to check if a sum less
// than pre_sum is present
public static int
b_search(int pre_sum, Dictionary<int, List<int> > m,
int index)
{
// Returns an iterator either equal
// to given key or next greater
int it = lowerBound(m, pre_sum);
if (it == 0)
return -1;
// Decrement the iterator
it--;
List<int> map_list = new List<int>();
map_list = m[it];
// Check if the sum found at
// a greater index
int it1 = lowerBoundList(map_list, index);
if (map_list[it1] > index)
return map_list[it1];
return -1;
}
// Function to return the index
// of first negative subarray sum
public static List<int> findSubarray(int[] arr, int n)
{
// Stores the prefix sum- index
// mappings
Dictionary<int, List<int> > m
= new Dictionary<int, List<int> >();
for(int i=0;i<n;i++)
{
List<int> a = new List<int>();
a.Add(0);
m.Add(i,a);
}
int sum = 0;
// Stores the prefix sum of
// the original array
int[] prefix_sum = new int[n];
for (int i = 0; i < n; i++) {
sum += arr[i];
// Check if we get sum negative
// starting from first element
if (sum < 0) {
List<int> xyz = new List<int>();
xyz.Add(0);
xyz.Add(i);
return xyz;
// return { 0, i };
}
prefix_sum[i] = sum;
m[sum].Add(i);
}
// Iterate through array find
// the sum which is just less
// then the previous prefix sum
for (int i = 1; i < n; i++) {
// Check if the starting element
// is itself negative
if (arr[i] < 0) {
// arr[i] becomes the required
// subarray
List<int> ret = new List<int>();
ret.Add(i);
ret.Add(i);
return ret;
// return { i, i };
}
else {
int pre_sum = prefix_sum[i - 1];
// Find the index which forms
// negative sum subarray
// from i-th index
int index = b_search(pre_sum, m, i);
// If a subarray is found
// starting from i-th index
if (index != -1) {
List<int> ret = new List<int>();
ret.Add(i);
ret.Add(index);
return ret;
// return { i, index };
}
}
}
// Return -1 if no such
// subarray is present
List<int> re = new List<int>();
re.Add(-1);
return re;
}
static public void Main()
{
// Given array arr[]
int[] arr = { 1, 2, -1, 3, -4, 3, -5 };
int n = arr.Length;
// Function Call
List<int> res = new List<int>();
res = findSubarray(arr, n);
// If subarray does not exist
if (res[0] == -1)
Console.WriteLine("-1");
// If the subarray exists
else {
Console.WriteLine(res[0] + " " + res[1]);
}
}
}
// This code is contributed by akashish__
<script>
// lower bound for a dictionary's key
function lowerBound(m, pre_sum)
{
let ans = -1;
for (const [key, value] of Object.entries(m)) {
if (key >= pre_sum) {
ans = key;
break;
}
}
return ans;
}
// lower bound for a list
function lowerBoundList(li, target)
{
let ans = -1;
for (let i = 0; i < li.Count; i++) {
if (li[i] >= target) {
ans = i;
break;
}
}
return ans;
}
// Function to check if a sum less
// than pre_sum is present
function b_search(pre_sum, m, index)
{
// Returns an iterator either equal
// to given key or next greater
let it = lowerBound(m, pre_sum);
if (it == 0)
return -1;
// Decrement the iterator
it--;
map_list = [];
map_list = m[it];
// Check if the sum found at
// a greater index
let it1 = lowerBoundList(map_list, index);
if (map_list[it1] > index)
return map_list[it1];
return -1;
}
// Function to return the index
// of first negative subarray sum
function findSubarray(/*int[]*/ arr, n)
{
// Stores the prefix sum- index
// mappings
m = {};
for(let i=0;i<n;i++)
{
a = [0];
m[i]=a;
}
let sum = 0;
// Stores the prefix sum of
// the original array
let prefix_sum = new Array(n);
for (let i = 0; i < n; i++) {
sum += arr[i];
// Check if we get sum negative
// starting from first element
if (sum < 0) {
xyz = [0,i];
return xyz;
}
prefix_sum[i] = sum;
m[sum] = i;
}
// Iterate through array find
// the sum which is just less
// then the previous prefix sum
for (let i = 1; i < n; i++) {
// Check if the starting element
// is itself negative
if (arr[i] < 0) {
// arr[i] becomes the required
// subarray
ret = [i,i];
return ret;
}
else {
let pre_sum = prefix_sum[i - 1];
// Find the index which forms
// negative sum subarray
// from i-th index
let index = b_search(pre_sum, m, i);
// If a subarray is found
// starting from i-th index
if (index != -1) {
ret = [i,index];
return ret;
}
}
}
// Return -1 if no such
// subarray is present
re = [-1];
return re;
}
// Given array arr[]
let arr = [ 1, 2, -1, 3, -4, 3, -5 ];
let n = arr.length;
// Function Call
let res = findSubarray(arr, n);
// If subarray does not exist
if (res[0] == -1)
console.log("-1");
// If the subarray exists
else {
console.log(res);
}
// This code is contributed by akashish__
</script>
Time Complexity: O(N * log N) Auxiliary Space: O(N)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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Given an array arr[] of either entirely positive integers or entirely negative integers except for one number. The task is to find that number. Examples: Input: arr[] = {3, 5, 2, 8, -7, 6, 9}Output: -7Explanation: Except -7 all the numbers in arr[] are positive integers. Input: arr[] = {-3, 5, -9}Ou
9 min read
Find the Sub-array with sum closest to 0
Given an array of both positive and negative numbers, the task is to find out the subarray whose sum is closest to 0. There can be multiple such subarrays, we need to output just 1 of them. Examples: Input : arr[] = {-1, 3, 2, -5, 4} Output : 1, 3 Subarray from index 1 to 3 has sum closest to 0 i.e.
13 min read