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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 <bits/stdc++.h>
using namespace std;
 
// Function to find the partition
void 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 partition
def 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 Code
if __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 partition
function 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>
Output: 
4.5

 

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


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