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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