Given an array of N integers, the task is to print the sum of the first subarray by splitting the array into exactly three subarrays such that the sum of the first and third subarray elements are equal and the maximum.
Note: All the elements must belong to a subarray and the subarrays can also be empty.
Examples:
Input: a[] = {1, 3, 1, 1, 4}
Output: 5
Split the N numbers to [1, 3, 1], [] and [1, 4]Input: a[] = {1, 3, 2, 1, 4}
Output: 4
Split the N numbers to [1, 3], [2, 1] and [4]
METHOD 1
A naive approach is to check for all possible partitions and use the prefix-sum concept to find out the partitions. The partition which gives the maximum sum of the first subarray will be the answer.
An efficient approach is as follows:
- Store the prefix sum and suffix sum of the N numbers.
- Hash the suffix sum’s index using a unordered_map in C++ or Hash-map in Java.
- Iterate from the beginning of the array, and check if the prefix sum exists in the suffix array beyond the current index i.
- If it does, then check for the previous maximum value and update accordingly.
Below is the implementation of the above approach:
C++
// C++ program for Split the array into three// subarrays such that summation of first// and third subarray is equal and maximum#include <bits/stdc++.h>using namespace std;// Function to return the sum of// the first subarrayint sumFirst(int a[], int n){ unordered_map<int, int> mp; int suf = 0; // calculate the suffix sum for (int i = n - 1; i >= 0; i--) { suf += a[i]; mp[suf] = i; } int pre = 0; int maxi = -1; // iterate from beginning for (int i = 0; i < n; i++) { // prefix sum pre += a[i]; // check if it exists beyond i if (mp[pre] > i) { // if greater then previous // then update maximum if (pre > maxi) { maxi = pre; } } } // First and second subarray empty if (maxi == -1) return 0; // partition done else return maxi;}// Driver Codeint main(){ int a[] = { 1, 3, 2, 1, 4 }; int n = sizeof(a) / sizeof(a[0]); // Function call cout << sumFirst(a, n); return 0;} |
Java
// Java program for Split the array into three// subarrays such that summation of first// and third subarray is equal and maximumimport java.util.HashMap;import java.util.Map;class GfG { // Function to return the sum // of the first subarray static int sumFirst(int a[], int n) { HashMap<Integer, Integer> mp = new HashMap<>(); int suf = 0; // calculate the suffix sum for (int i = n - 1; i >= 0; i--) { suf += a[i]; mp.put(suf, i); } int pre = 0, maxi = -1; // iterate from beginning for (int i = 0; i < n; i++) { // prefix sum pre += a[i]; // check if it exists beyond i if (mp.containsKey(pre) && mp.get(pre) > i) { // if greater then previous // then update maximum if (pre > maxi) { maxi = pre; } } } // First and second subarray empty if (maxi == -1) return 0; // partition done else return maxi; } // Driver code public static void main(String[] args) { int a[] = { 1, 3, 2, 1, 4 }; int n = a.length; // Function call System.out.println(sumFirst(a, n)); }}// This code is contributed by Rituraj Jain |
Python3
# Python3 program for Split the array into three# subarrays such that summation of first# and third subarray is equal and maximum# Function to return the sum of# the first subarraydef sumFirst(a, n): mp = {i: 0 for i in range(7)} suf = 0 i = n - 1 # calculate the suffix sum while(i >= 0): suf += a[i] mp[suf] = i i -= 1 pre = 0 maxi = -1 # iterate from beginning for i in range(n): # prefix sum pre += a[i] # check if it exists beyond i if (mp[pre] > i): # if greater then previous # then update maximum if (pre > maxi): maxi = pre # First and second subarray empty if (maxi == -1): return 0 # partition done else: return maxi# Driver Codeif __name__ == '__main__': a = [1, 3, 2, 1, 4] n = len(a) # Function call print(sumFirst(a, n))# This code is contributed by# Surendra_Gangwar |
C#
// C# program for Split the array into three// subarrays such that summation of first// and third subarray is equal and maximumusing System;using System.Collections.Generic;class GfG { // Function to return the sum // of the first subarray static int sumFirst(int[] a, int n) { Dictionary<int, int> mp = new Dictionary<int, int>(); int suf = 0; // calculate the suffix sum for (int i = n - 1; i >= 0; i--) { suf += a[i]; mp.Add(suf, i); if (mp.ContainsKey(suf)) { mp.Remove(suf); } mp.Add(suf, i); } int pre = 0, maxi = -1; // iterate from beginning for (int i = 0; i < n; i++) { // prefix sum pre += a[i]; // check if it exists beyond i if (mp.ContainsKey(pre) && mp[pre] > i) { // if greater then previous // then update maximum if (pre > maxi) { maxi = pre; } } } // First and second subarray empty if (maxi == -1) return 0; // partition done else return maxi; } // Driver code public static void Main(String[] args) { int[] a = { 1, 3, 2, 1, 4 }; int n = a.Length; // Function call Console.WriteLine(sumFirst(a, n)); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript program for Split the array into three// subarrays such that summation of first// and third subarray is equal and maximum// Function to return the sum of// the first subarrayfunction sumFirst(a, n){ var mp = new Map(); var suf = 0; // calculate the suffix sum for (var i = n - 1; i >= 0; i--) { suf += a[i]; mp.set(suf, i); } var pre = 0; var maxi = -1; // iterate from beginning for (var i = 0; i < n; i++) { // prefix sum pre += a[i]; // check if it exists beyond i if (mp.get(pre) > i) { // if greater then previous // then update maximum if (pre > maxi) { maxi = pre; } } } // First and second subarray empty if (maxi == -1) return 0; // partition done else return maxi;}// Driver Codevar a = [1, 3, 2, 1, 4];var n = a.length;// Function calldocument.write( sumFirst(a, n));</script> |
Output
4
Time Complexity: O(n) where n is the size of the given array
Auxiliary Space: O(n)
METHOD 2
Approach: We will use two pointers concept where one pointer will start from the front and the other from the back. In each iteration, the sum of the first and last subarray is compared and if they are the same then the sum is updated in the answer variable.
Algorithm:
- Initialize front_pointer to 0 and back_pointer to n-1.
- Initialize prefixsum to arr[ front_pointer ] and suffixsum to arr[back_pointer].
- The summations are compared.
- If prefixsum > suffixsum ,back_pointer is decremented by 1 and suffixsum+= arr[ back_pointer ].
- If prefixsum < suffixsum, front_pointer is incremented by 1 and prefixsum+= arr[ front_pointer ]
- If they are the same then the sum is updated in the answer variable and both the pointers are moved by one step and both prefixsum and suffixsum are updated accordingly.
- The above step is continued until the front_pointer is no less than the back_pointer.
Below is the implementation of the above approach:
C++
// C++ program for Split the array into three// subarrays such that summation of first// and third subarray is equal and maximum#include <bits/stdc++.h>using namespace std;// Function to return the sum of// the first subarrayint sumFirst(int a[], int n){ // two pointers are initialized // one at the front and the other // at the back int front_pointer = 0; int back_pointer = n - 1; // prefixsum and suffixsum initialized int prefixsum = a[front_pointer]; int suffixsum = a[back_pointer]; // answer variable initialized to 0 int answer = 0; while (front_pointer < back_pointer) { // if the summation are equal if (prefixsum == suffixsum) { // answer updated answer = max(answer, prefixsum); // both the pointers are moved by step front_pointer++; back_pointer--; // prefixsum and suffixsum are updated prefixsum += a[front_pointer]; suffixsum += a[back_pointer]; } else if (prefixsum > suffixsum) { // if prefixsum is more,then back pointer is // moved by one step and suffixsum updated. back_pointer--; suffixsum += a[back_pointer]; } else { // if prefixsum is less,then front pointer is // moved by one step and prefixsum updated. front_pointer++; prefixsum += a[front_pointer]; } } // answer is returned return answer;}// Driver codeint main(){ int arr[] = { 1, 3, 2, 1, 4 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call cout << sumFirst(arr, n); // This code is contributed by Arif} |
Java
// Java program for Split the array into three// subarrays such that summation of first// and third subarray is equal and maximumimport java.util.*;class GFG{ // Function to return the sum of// the first subarraypublic static int sumFirst(int a[], int n){ // Two pointers are initialized // one at the front and the other // at the back int front_pointer = 0; int back_pointer = n - 1; // prefixsum and suffixsum initialized int prefixsum = a[front_pointer]; int suffixsum = a[back_pointer]; // answer variable initialized to 0 int answer = 0; while (front_pointer < back_pointer) { // If the summation are equal if (prefixsum == suffixsum) { // answer updated answer = Math.max(answer, prefixsum); // Both the pointers are moved by step front_pointer++; back_pointer--; // prefixsum and suffixsum are updated prefixsum += a[front_pointer]; suffixsum += a[back_pointer]; } else if (prefixsum > suffixsum) { // If prefixsum is more,then back pointer is // moved by one step and suffixsum updated. back_pointer--; suffixsum += a[back_pointer]; } else { // If prefixsum is less,then front pointer is // moved by one step and prefixsum updated. front_pointer++; prefixsum += a[front_pointer]; } } // answer is returned return answer;} // Driver codepublic static void main(String[] args){ int arr[] = { 1, 3, 2, 1, 4 }; int n = arr.length; // Function call System.out.print(sumFirst(arr, n));}}// This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program for Split the array into three# subarrays such that summation of first# and third subarray is equal and maximumimport math # Function to return the sum of# the first subarraydef sumFirst(a, n): # Two pointers are initialized # one at the front and the other # at the back front_pointer = 0 back_pointer = n - 1 # prefixsum and suffixsum initialized prefixsum = a[front_pointer] suffixsum = a[back_pointer] # answer variable initialized to 0 answer = 0 while (front_pointer < back_pointer): # If the summation are equal if (prefixsum == suffixsum): # answer updated answer = max(answer, prefixsum) # Both the pointers are moved by step front_pointer += 1 back_pointer -= 1 # prefixsum and suffixsum are updated prefixsum += a[front_pointer] suffixsum += a[back_pointer] elif (prefixsum > suffixsum): # If prefixsum is more,then back pointer is # moved by one step and suffixsum updated. back_pointer -= 1 suffixsum += a[back_pointer] else: # If prefixsum is less,then front pointer is # moved by one step and prefixsum updated. front_pointer += 1 prefixsum += a[front_pointer] # answer is returned return answer# Driver codearr = [ 1, 3, 2, 1, 4 ]n = len(arr)# Function callprint(sumFirst(arr, n))# This code is contributed by Stream_Cipher |
C#
// C# program for split the array into three// subarrays such that summation of first// and third subarray is equal and maximumusing System;class GFG{ // Function to return the sum of// the first subarraystatic int sumFirst(int[] a, int n){ // Two pointers are initialized // one at the front and the other // at the back int front_pointer = 0; int back_pointer = n - 1; // prefixsum and suffixsum initialized int prefixsum = a[front_pointer]; int suffixsum = a[back_pointer]; // answer variable initialized to 0 int answer = 0; while (front_pointer < back_pointer) { // If the summation are equal if (prefixsum == suffixsum) { // answer updated answer = Math.Max(answer, prefixsum); // Both the pointers are moved by step front_pointer++; back_pointer--; // prefixsum and suffixsum are updated prefixsum += a[front_pointer]; suffixsum += a[back_pointer]; } else if (prefixsum > suffixsum) { // If prefixsum is more,then back pointer is // moved by one step and suffixsum updated. back_pointer--; suffixsum += a[back_pointer]; } else { // If prefixsum is less,then front pointer is // moved by one step and prefixsum updated. front_pointer++; prefixsum += a[front_pointer]; } } // answer is returned return answer;} // Driver Codestatic void Main() { int[] arr = { 1, 3, 2, 1, 4 }; int n = arr.Length; // Function call Console.WriteLine(sumFirst(arr, n));}}// This code is contributed by divyesh072019 |
Javascript
<script>// javascript program for Split the array into three// subarrays such that summation of first// and third subarray is equal and maximum // Function to return the sum of // the first subarray function sumFirst(a, n) { // Two pointers are initialized // one at the front and the other // at the back var front_pointer = 0; var back_pointer = n - 1; // prefixsum and suffixsum initialized var prefixsum = a[front_pointer]; var suffixsum = a[back_pointer]; // answer variable initialized to 0 var answer = 0; while (front_pointer < back_pointer) { // If the summation are equal if (prefixsum == suffixsum) { // answer updated answer = Math.max(answer, prefixsum); // Both the pointers are moved by step front_pointer++; back_pointer--; // prefixsum and suffixsum are updated prefixsum += a[front_pointer]; suffixsum += a[back_pointer]; } else if (prefixsum > suffixsum) { // If prefixsum is more,then back pointer is // moved by one step and suffixsum updated. back_pointer--; suffixsum += a[back_pointer]; } else { // If prefixsum is less,then front pointer is // moved by one step and prefixsum updated. front_pointer++; prefixsum += a[front_pointer]; } } // answer is returned return answer; } // Driver code var arr = [ 1, 3, 2, 1, 4 ]; var n = arr.length; // Function call document.write(sumFirst(arr, n));// This code is contributed by todaysgaurav</script> |
Output
4
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
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