Sudo Placement[1.5] | Partition

• Difficulty Level : Medium
• Last Updated : 16 Jun, 2021

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:
Left array = {1, 2, -1, 2}
Right array = {3}
Sumleft = 4, Sumright = 3
Difference = 1 which is the minimum possible
Input: 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 using namespace std; // Function to find the partitionvoid partition(int a[], int n){    unordered_map 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- partitionimport java.util.*; class GFG{     // Function to find the partition    static void partition(int a[], int n)    {        Map 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- partitionusing System;using System.Collections.Generic; class GFG{     // Function to find the partition    static void partition(int []a, int n)    {        Dictionary mpp = new Dictionary();         // 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


Output:
4.5

Time Complexity: O(n) under the assumption that unordered_map search works in O(1) time.

My Personal Notes arrow_drop_up