Given an array of positive and negative numbers. The task is to find a partition point such that none of the elements of left array are in the right array. If there are multiple partitions, then find the partition at which the absolute difference between the sum of left array and sum of right array (|sumleft – sumright|) with respect to the partition point is minimum. In case of multiple points, print the first partition point from left which is (last index of left array and first index of right array)/2 . Consider 1-based indexing. The left and right array on partition must have a minimum of 1 element and maximum of n-1 elements. Print -1 if no partition is possible.
Examples:
Input: a[] = {1, 2, -1, 2, 3}
Output: 1
Left array = {1, 2, -1, 2}
Right array = {3}
Sumleft = 4, Sumright = 3
Difference = 1 which is the minimum possibleInput: a[] = {1, 2, 3, 1}
Output: -1
A naive approach will be to traverse left and right from every index and check if the partition is possible or not at that index. If the partition is possible, then check if the absolute difference between the sum of an element of left array and element of right array is less than that of the previous obtained value at the partition. After finding the partition point, greedily find the |sumleft – sumright|.
Time Complexity: O(N2)
An efficient solution will be to store the last index of every occurring element in a hash-map. Since the element values are large, direct indexing cannot be used. Create a prefix[] and suffix[] array which stores the prefix sum and suffix sum respectively. Initialize a variable count as 0. Iterate for all the element in the array. A common point of observation is, while traversing if the present element’s(Ai) last nonoccurence is not i itself, then we cannot have a partition in between i and the element’s last occurrence. While traversing store the maximum of element’s last occurrence as the partition cannot be done till then.
Once the count is i itself, we can have a partition, now if there are multiple partitions then choose the min |sumleft – sumright|.
Note : Use of map instead of unordered_map may cause TLE.
Below is the implementation of the above approach.
C++
// C++ program for SP- partition#include <bits/stdc++.h>using namespace std;// Function to find the partitionvoid partition(int a[], int n){ unordered_map<long long, long long> mpp; // mark the last occurrence of every element for (int i = 0; i < n; i++) mpp[a[i]] = i; // calculate the prefix sum long long presum[n]; presum[0] = a[0]; for (int i = 1; i < n; i++) presum[i] = presum[i - 1] + a[i]; // calculate the suffix sum long long sufsum[n]; sufsum[n - 1] = a[n - 1]; for (int i = n - 2; i >= 0; i--) { sufsum[i] = sufsum[i + 1] + a[i]; } // Check if partition is possible bool possible = false; // Stores the absolute difference long long ans = 1e18; // stores the last index till // which there can not be any partition long long count = 0; // Stores the partition long long index = -1; // Check if partition is possible or not // donot check for the last element // as partition is not possible for (int i = 0; i < n - 1; i++) { // takes an element and checks it last occurrence // stores the maximum of the last occurrence // where partition can be done count = max(count, mpp[a[i]]); // if partition is possible if (count == i) { // partition is possible possible = true; // stores the left array sum long long sumleft = presum[i]; // stores the right array sum long long sumright = sufsum[i + 1]; // check if the difference is minimum if ((abs(sumleft - sumright)) < ans) { ans = abs(sumleft - sumright); index = i + 1; } } } // is partition is possible or not if (possible) cout << index << ".5" << endl; else cout << -1 << endl;}// Driver Code-int main(){ int a[] = { 1, 2, -1, 2, 3 }; int n = sizeof(a) / sizeof(a[0]); partition(a, n); return 0;} |
Java
// Java program for SP- partition import java.util.*;class GFG { // Function to find the partition static void partition(int a[], int n) { Map<Integer, Integer> mpp = new HashMap<>(); // mark the last occurrence of // every element for (int i = 0; i < n; i++) mpp.put(a[i], i); // calculate the prefix sum long[] presum = new long[n]; presum[0] = a[0]; for (int i = 1; i < n; i++) presum[i] = presum[i - 1] + a[i]; // calculate the suffix sum long[] sufsum = new long[n]; sufsum[n - 1] = a[n - 1]; for (int i = n - 2; i >= 0; i--) { sufsum[i] = sufsum[i + 1] + a[i]; } // Check if partition is possible boolean possible = false; // Stores the absolute difference long ans = (long) 1e18; // stores the last index till // which there can not be any partition long count = 0; // Stores the partition long index = -1; // Check if partition is possible or not // donot check for the last element // as partition is not possible for (int i = 0; i < n - 1; i++) { // takes an element and checks its // last occurrence, stores the maximum // of the last occurrence where // partition can be done count = Math.max(count, mpp.get(a[i])); // if partition is possible if (count == i) { // partition is possible possible = true; // stores the left array sum long sumleft = presum[i]; // stores the right array sum long sumright = sufsum[i + 1]; // check if the difference is minimum if ((Math.abs(sumleft - sumright)) < ans) { ans = Math.abs(sumleft - sumright); index = i + 1; } } } // is partition is possible or not if (possible) System.out.print(index + ".5" + "\n"); else System.out.print(-1 + "\n"); } // Driver Code public static void main(String[] args) { int a[] = { 1, 2, -1, 2, 3 }; int n = a.length; partition(a, n); }}// This code is contributed by 29AjayKumar |
Python3
# Python program for SP- partition# Function to find the partitiondef partition(a: list, n: int): mpp = dict() # mark the last occurrence of every element for i in range(n): mpp[a[i]] = i # calculate the prefix sum preSum = [0] * n preSum[0] = a[0] for i in range(1, n): preSum[i] = preSum[i - 1] + a[i] # calculate the suffix sum sufSum = [0] * n sufSum[n - 1] = a[n - 1] for i in range(n - 2, -1, -1): sufSum[i] = sufSum[i + 1] + a[i] # Check if partition is possible possible = False # Stores the absolute difference ans = int(1e18) # stores the last index till # which there can not be any partition count = 0 # Stores the partition index = -1 # Check if partition is possible or not # donot check for the last element # as partition is not possible for i in range(n - 1): # takes an element and checks it last occurrence # stores the maximum of the last occurrence # where partition can be done count = max(count, mpp[a[i]]) # if partition is possible if count == i: # partition is possible possible = True # stores the left array sum sumleft = preSum[i] # stores the right array sum sumright = sufSum[i + 1] # check if the difference is minimum if abs(sumleft - sumright) < ans: ans = abs(sumleft - sumright) index = i + 1 # is partition is possible or not if possible: print("%d.5" % index) else: print("-1")# Driver Codeif __name__ == "__main__": a = [1, 2, -1, 2, 3] n = len(a) partition(a, n)# This code is contributed by# sanjeev2552 |
C#
// C# program for SP- partition using System;using System.Collections.Generic;class GFG { // Function to find the partition static void partition(int []a, int n) { Dictionary<int, int> mpp = new Dictionary<int, int>(); // mark the last occurrence of // every element for (int i = 0; i < n; i++) if(mpp.ContainsKey(a[i])) mpp[a[i]] = i; else mpp.Add(a[i], i); // calculate the prefix sum long[] presum = new long[n]; presum[0] = a[0]; for (int i = 1; i < n; i++) presum[i] = presum[i - 1] + a[i]; // calculate the suffix sum long[] sufsum = new long[n]; sufsum[n - 1] = a[n - 1]; for (int i = n - 2; i >= 0; i--) { sufsum[i] = sufsum[i + 1] + a[i]; } // Check if partition is possible bool possible = false; // Stores the absolute difference long ans = (long) 1e18; // stores the last index till which // there can not be any partition long count = 0; // Stores the partition long index = -1; // Check if partition is possible or not // donot check for the last element // as partition is not possible for (int i = 0; i < n - 1; i++) { // takes an element and checks its // last occurrence, stores the maximum // of the last occurrence where // partition can be done count = Math.Max(count, mpp[a[i]]); // if partition is possible if (count == i) { // partition is possible possible = true; // stores the left array sum long sumleft = presum[i]; // stores the right array sum long sumright = sufsum[i + 1]; // check if the difference is minimum if ((Math.Abs(sumleft - sumright)) < ans) { ans = Math.Abs(sumleft - sumright); index = i + 1; } } } // is partition is possible or not if (possible) Console.Write(index + ".5" + "\n"); else Console.Write(-1 + "\n"); } // Driver Code public static void Main(String[] args) { int []a = { 1, 2, -1, 2, 3 }; int n = a.Length; partition(a, n); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program for SP- partition// Function to find the partitionfunction partition(a, n) { let mpp = new Map(); // mark the last occurrence of every element for (let i = 0; i < n; i++) mpp.set(a[i], i); // calculate the prefix sum let presum = new Array(n); presum[0] = a[0]; for (let i = 1; i < n; i++) presum[i] = presum[i - 1] + a[i]; // calculate the suffix sum let sufsum = new Array(n); sufsum[n - 1] = a[n - 1]; for (let i = n - 2; i >= 0; i--) { sufsum[i] = sufsum[i + 1] + a[i]; } // Check if partition is possible let possible = false; // Stores the absolute difference let ans = Number.MAX_SAFE_INTEGER; // stores the last index till // which there can not be any partition let count = 0; // Stores the partition let index = -1; // Check if partition is possible or not // donot check for the last element // as partition is not possible for (let i = 0; i < n - 1; i++) { // takes an element and checks it last occurrence // stores the maximum of the last occurrence // where partition can be done count = Math.max(count, mpp.get(a[i])); // if partition is possible if (count == i) { // partition is possible possible = true; // stores the left array sum let sumleft = presum[i]; // stores the right array sum let sumright = sufsum[i + 1]; // check if the difference is minimum if ((Math.abs(sumleft - sumright)) < ans) { ans = Math.abs(sumleft - sumright); index = i + 1; } } } // is partition is possible or not if (possible) document.write(index + ".5" + "<br>"); else document.write(-1 + "<br>");}// Driver Code-let a = [1, 2, -1, 2, 3];let n = a.length;partition(a, n);// This code is contributed by saurabh_jaiswal.</script> |
4.5
Time Complexity: O(n) under the assumption that unordered_map search works in O(1) time.
Auxiliary Space: O(n)
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