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 sumvector<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 Codeint 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;} |
Java
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)); } }} |
Python3
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 Codeif __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]) |
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 presentint 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 sumvector<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 Codeint 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;} |
Java
/*package whatever //do not write package name here */// Java program for the above approachimport 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__ |
Python3
# 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 listdef 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 presentdef 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 sumdef 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 Callres = findSubarray(arr, n)# If subarray does not existif (res[0] == -1): print("-1")# If the subarray existselse: print(res)# This code is contributed by akashish__ |
C#
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__ |
Javascript
<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 listfunction 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 presentfunction 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 sumfunction 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 Calllet res = findSubarray(arr, n);// If subarray does not existif (res[0] == -1) console.log("-1");// If the subarray existselse { console.log(res);}// This code is contributed by akashish__</script> |
Output
0 6
Time Complexity: O(N * log N) Auxiliary Space: O(N)
Related Topic: Subarrays, Subsequences, and Subsets in Array
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
