Optimal partition of an array into four parts
Last Updated :
19 Oct, 2022
Given an array of n non-negative integers. Choose three indices i.e. (0 <= index_1 <= index_ 2<= index_3 <= n) from the array to make four subsets such that the term sum(0, index_1) – sum(index_1, index_2) + sum(index_2, index_3) – sum(index_3, n) is maximum possible.
Here, two indices say l and r means, the sum(l, r) will be the sum of all numbers of subset on positions from l to r non-inclusive (l-th element is not counted, r-th element is counted).
For example, if arr = {-5, 3, 9, 4}, then sum(0, 1) = -5, sum(0, 2) = -2, sum(1, 4) = 16 and sum(i, i) = 0 for each i from 0 to 4. For indices l and r holds 0 <= l <= r <= n. Indices in array are numbered from 0.
Examples:
Input : arr = {-1, 2, 3}
Output : 0 1 3
Here, sum(0, 0) = 0
sum(0, 1) = -1
sum(1, 3) = 2 + 3 = 5
sum(3, 3) = 0
Therefore , sum(0, 0) - sum(0, 1) + sum(1, 3) - sum(3, 3) = 4
which is maximum.
Input : arr = {0, 0, -1, 0}
Output : 0 0 0
Here, sum(0, 0) - sum(0, 0) + sum(0, 0) - sum(0, 0) = 0
which is maximum possible.
Imagine the same task but without the first term in sum. As the sum of the array is fixed, the best second segment should be the one with the greatest sum. This can be solved in O(n) with prefix sum. When recalling the best segment to end at position i, take minimal prefix sum from 0 to i inclusive (from the whole sum you want to subtract the lowest number).
Now let’s just iterate over all possible ends of the first segment and solve the task above on the array without this segment.
Implementation:
C++
#include <bits/stdc++.h>
#define max 50009
using namespace std;
void find_Indices( int arr[], int n){
int sum[max], k;
int index_1, index_2, index_3, index;
for ( int i = 1, k = 0; i <= n; i++)
sum[i] = sum[i-1] + arr[k++];
long long ans = -(1e15);
index_1 = index_2 = index_3 = -1;
for ( int l = 0; l <= n; l++) {
int index = 0;
long long vmin = (1e15);
for ( int r = l; r <= n; r++) {
if (sum[r] < vmin) {
vmin = sum[r];
index = r;
}
if (sum[l] + sum[r] - vmin > ans)
{
ans = sum[l] + sum[r] - vmin;
index_1 = l;
index_2 = index;
index_3 = r;
}
}
}
printf ( "%d %d %d" , index_1, index_2, index_3);
}
int main() {
int arr[] = {-1, 2, 3};
int n = sizeof (arr)/ sizeof (arr[0]);
find_Indices(arr, n);
}
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Java
class GFG {
static final int max = 50009 ;
static void find_Indices( int arr[], int n)
{
int sum[] = new int [max];
int index_1, index_2, index_3, index;
int k, i;
for (i = 1 , k = 0 ; i <= n; i++)
sum[i] = sum[i - 1 ] + arr[k++];
double ans = -(1e15);
index_1 = index_2 = index_3 = - 1 ;
for ( int l = 0 ; l <= n; l++) {
index = 0 ;
double vmin = (1e15);
for ( int r = l; r <= n; r++) {
if (sum[r] < vmin)
{
vmin = sum[r];
index = r;
}
if (sum[l] + sum[r] - vmin > ans)
{
ans = sum[l] + sum[r] - vmin;
index_1 = l;
index_2 = index;
index_3 = r;
}
}
}
System.out.print(index_1 + " " + index_2 +
" " + index_3);
}
public static void main(String[] args)
{
int arr[] = { - 1 , 2 , 3 };
int n = arr.length;
find_Indices(arr, n);
}
}
|
Python3
max = 50009
def find_Indices(arr, n):
sum = [ 0 for i in range ( max )]
k = 0
for i in range ( 1 ,n + 1 ):
sum [i] = sum [i - 1 ] + arr[k];
k + = 1
ans = - ( 1e15 )
index_1 = index_2 = index_3 = - 1
for l in range (n + 1 ):
index = 0
vmin = ( 1e15 )
for r in range (l,n + 1 ):
if ( sum [r] < vmin):
vmin = sum [r]
index = r
if ( sum [l] + sum [r] - vmin > ans):
ans = sum [l] + sum [r] - vmin
index_1 = l
index_2 = index
index_3 = r
print (index_1, " " , index_2, " " , index_3)
arr = [ - 1 , 2 , 3 ]
n = len (arr)
find_Indices(arr, n)
|
C#
using System;
class GFG {
static int max = 50009;
static void find_Indices( int []arr, int n)
{
int []sum = new int [max];
int index_1, index_2, index_3, index;
int k, i;
for (i = 1, k = 0; i <= n; i++)
sum[i] = sum[i - 1] + arr[k++];
double ans = -(1e15);
index_1 = index_2 = index_3 = -1;
for ( int l = 0; l <= n; l++) {
index = 0;
double vmin = (1e15);
for ( int r = l; r <= n; r++) {
if (sum[r] < vmin)
{
vmin = sum[r];
index = r;
}
if (sum[l] + sum[r] - vmin > ans)
{
ans = sum[l] + sum[r] - vmin;
index_1 = l;
index_2 = index;
index_3 = r;
}
}
}
Console.WriteLine(index_1 + " " + index_2 +
" " + index_3);
}
public static void Main()
{
int []arr = { -1, 2, 3 };
int n = arr.Length;
find_Indices(arr, n);
}
}
|
PHP
<?php
$max = 50009;
function find_Indices( $arr , $n )
{
global $max ;
$sum = array (); $k = 0;
$sum [0] = 0;
for ( $i = 1, $k = 0;
$i <= $n ; $i ++)
$sum [ $i ] = $sum [ $i - 1] +
$arr [ $k ++];
$ans = -(1000000000000000);
$index_1 = $index_2 = $index_3 = -1;
for ( $l = 0; $l <= $n ; $l ++)
{
$index = 0;
$vmin = (1000000000000000);
for ( $r = $l ; $r <= $n ; $r ++)
{
if ( $sum [ $r ] < $vmin )
{
$vmin = $sum [ $r ];
$index = $r ;
}
if ( $sum [ $l ] + $sum [ $r ] -
$vmin > $ans )
{
$ans = $sum [ $l ] +
$sum [ $r ] - $vmin ;
$index_1 = $l ;
$index_2 = $index ;
$index_3 = $r ;
}
}
}
echo ( $index_1 . " " . $index_2 .
" " . $index_3 . " " );
}
$arr = array (-1, 2, 3);
$n = count ( $arr );
find_Indices( $arr , $n );
?>
|
Javascript
<script>
let max = 50009;
function find_Indices(arr, n)
{
let sum = new Array(); k = 0;
sum[0] = 0;
for (let i = 1, k = 0;
i <= n; i++)
sum[i] = sum[i - 1] +
arr[k++];
let ans = -(1000000000000000);
let index_1 = index_2 = index_3 = -1;
for (let l = 0; l <= n; l++)
{
let index = 0;
let vmin = (1000000000000000);
for (let r = l; r <= n; r++)
{
if (sum[r] < vmin)
{
vmin = sum[r];
index = r;
}
if (sum[l] + sum[r] -
vmin > ans)
{
ans = sum[l] +
sum[r] - vmin;
index_1 = l;
index_2 = index;
index_3 = r;
}
}
}
document.write(index_1 + " " + index_2 + " " + index_3 + " " );
}
let arr = new Array(-1, 2, 3);
let n = arr.length;
find_Indices(arr, n);
</script>
|
Time complexity:
Auxiliary space: O(max) as it is using extra space for the array sum
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