Optimal partition of an array into four parts

• Last Updated : 01 Aug, 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++

 `// CPP program to find three indices` `#include ` `#define max 50009` `using` `namespace` `std;`   `// Function to find required indices.` `void` `find_Indices(``int` `arr[], ``int` `n){` `    ``int` `sum[max], k;` `    ``int` `index_1, index_2, index_3, index;` `    `  `    ``// calculating prefix sum from` `    ``// 1 to i for each i.` `    ``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;`   `    ``// iterating the loop from 0 to n` `    ``// for all possibilities.` `    ``for` `(``int` `l = 0; l <= n; l++) {` `        ``int` `index = 0;` `        ``long` `long` `vmin = (1e15);`   `        ``// Here, we recalling the best` `        ``// segment to end at position i.` `        ``for` `(``int` `r = l; r <= n; r++) {`   `            ``// taking the minimal prefix` `            ``// sum from 0 to i inclusive.` `            ``if` `(sum[r] < vmin) {` `                ``vmin = sum[r];` `                ``index = r;` `            ``}` `                        `  `            ``// calculating the indices.` `            ``if` `(sum[l] + sum[r] - vmin > ans) ` `            ``{` `                ``ans = sum[l] + sum[r] - vmin;` `                ``index_1 = l;` `                ``index_2 = index;` `                ``index_3 = r;` `            ``}` `        ``}` `    ``}`   `    ``// Required indices.` `    ``printf``(``"%d %d %d"``, index_1, index_2, index_3);` `}`   `// Driver function` `int` `main() {` `    ``int` `arr[] = {-1, 2, 3};` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);` `    ``find_Indices(arr, n);` `}`

Java

 `// Java program to find three indices` `class` `GFG {` `    `  `    ``static` `final` `int` `max = ``50009``;` `    `  `    ``// Function to find required indices.` `    ``static` `void` `find_Indices(``int` `arr[], ``int` `n)` `    ``{` `        `  `        ``int` `sum[] = ``new` `int``[max];` `        ``int` `index_1, index_2, index_3, index;` `        ``int` `k, i;`   `        ``// calculating prefix sum from` `        ``// 1 to i for each 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``;`   `        ``// iterating the loop from 0 to n` `        ``// for all possibilities.` `        ``for` `(``int` `l = ``0``; l <= n; l++) {` `            ``index = ``0``;` `            ``double` `vmin = (1e15);`   `            ``// Here, we recalling the best` `            ``// segment to end at position i.` `            ``for` `(``int` `r = l; r <= n; r++) {`   `                ``// taking the minimal prefix` `                ``// sum from 0 to i inclusive.` `                ``if` `(sum[r] < vmin) ` `                ``{` `                    ``vmin = sum[r];` `                    ``index = r;` `                ``}`   `                ``// calculating the indices.` `                ``if` `(sum[l] + sum[r] - vmin > ans) ` `                ``{` `                    ``ans = sum[l] + sum[r] - vmin;` `                    ``index_1 = l;` `                    ``index_2 = index;` `                    ``index_3 = r;` `                ``}` `            ``}` `        ``}`   `        ``// Required indices.` `        ``System.out.print(index_1 + ``" "` `+ index_2 +` `                                    ``" "` `+ index_3);` `    ``}` `    `  `    ``// Driver function.` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { -``1``, ``2``, ``3` `};` `        ``int` `n = arr.length;`   `        ``find_Indices(arr, n);` `    ``}` `}`   `// This code is contributed by Anant Agarwal.`

Python3

 `# Python program to` `# find three indices`   `max` `=` `50009`   `# Function to find` `# required indices.` `def` `find_Indices(arr, n):`   `    ``sum``=``[``0` `for` `i ``in` `range``(``max``)]` `     `  `    ``# calculating prefix sum from` `    ``# 1 to i for each i.` `    ``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` ` `  `    ``# iterating the loop from 0 to n` `    ``# for all possibilities.` `    ``for` `l ``in` `range``(n``+``1``):` `        ``index ``=` `0` `        ``vmin ``=` `(``1e15``)` ` `  `        ``# Here, we recalling the best` `        ``# segment to end at position i.` `        ``for` `r ``in` `range``(l,n``+``1``):` ` `  `            `  `            ``# taking the minimal prefix` `            ``# sum from 0 to i inclusive.` `            ``if` `(``sum``[r] < vmin):` `                ``vmin ``=` `sum``[r]` `                ``index ``=` `r` `            `  `                         `  `            ``# calculating the indices.` `            ``if` `(``sum``[l] ``+` `sum``[r] ``-` `vmin > ans): ` `            `  `                ``ans ``=` `sum``[l] ``+` `sum``[r] ``-` `vmin` `                ``index_1 ``=` `l` `                ``index_2 ``=` `index` `                ``index_3 ``=` `r` `        `  `    ``# Required indices.` `    ``print``(index_1,``" "``, index_2,``" "``, index_3)` ` `  `# Driver function`   `arr ``=` `[``-``1``, ``2``, ``3``]` `n ``=` `len``(arr)` `find_Indices(arr, n)`   `# This code is contributed` `# by Anant Agarwal.`

C#

 `// C# program to find three indices` `using` `System;`   `class` `GFG {` `    `  `    ``static` `int` `max = 50009;` `    `  `    ``// Function to find required indices.` `    ``static` `void` `find_Indices(``int` `[]arr, ``int` `n)` `    ``{` `        `  `        ``int` `[]sum = ``new` `int``[max];` `        ``int` `index_1, index_2, index_3, index;` `        ``int` `k, i;`   `        ``// calculating prefix sum from` `        ``// 1 to i for each 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;`   `        ``// iterating the loop from 0 to n` `        ``// for all possibilities.` `        ``for` `(``int` `l = 0; l <= n; l++) {` `            ``index = 0;` `            ``double` `vmin = (1e15);`   `            ``// Here, we recalling the best` `            ``// segment to end at position i.` `            ``for` `(``int` `r = l; r <= n; r++) {`   `                ``// taking the minimal prefix` `                ``// sum from 0 to i inclusive.` `                ``if` `(sum[r] < vmin) ` `                ``{` `                    ``vmin = sum[r];` `                    ``index = r;` `                ``}`   `                ``// calculating the indices.` `                ``if` `(sum[l] + sum[r] - vmin > ans) ` `                ``{` `                    ``ans = sum[l] + sum[r] - vmin;` `                    ``index_1 = l;` `                    ``index_2 = index;` `                    ``index_3 = r;` `                ``}` `            ``}` `        ``}`   `        ``// Required indices.` `        ``Console.WriteLine(index_1 + ``" "` `+ index_2 +` `                                    ``" "` `+ index_3);` `    ``}` `    `  `    ``// Driver function.` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `[]arr = { -1, 2, 3 };` `        ``int` `n = arr.Length;`   `        ``find_Indices(arr, n);` `    ``}` `}`   `// This code is contributed by vt_m.`

PHP

 ` ``\$ans``) ` `            ``{` `                ``\$ans` `= ``\$sum``[``\$l``] + ` `                       ``\$sum``[``\$r``] - ``\$vmin``;` `                ``\$index_1` `= ``\$l``;` `                ``\$index_2` `= ``\$index``;` `                ``\$index_3` `= ``\$r``;` `            ``}` `        ``}` `    ``}`   `    ``// Required indices.` `    ``echo``(``\$index_1``.``" "``.``\$index_2``.` `                  ``" "``.``\$index_3``.``" "``);` `}`   `// Driver Code` `\$arr` `= ``array``(-1, 2, 3);` `\$n` `= ``count``(``\$arr``);` `find_Indices(``\$arr``, ``\$n``);`   `// This code is contributed by` `// Manish Shaw(manishshaw1)` `?>`

Javascript

 ``

Output

`0 1 3`

Time complexity:

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