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++ 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 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 |
# 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# 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 |
<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> |
4.5
Time Complexity: O(n) under the assumption that unordered_map search works in O(1) time.
Auxiliary Space: O(n)