Open In App

Maximum sub-array sum after dividing array into sub-arrays based on the given queries

Improve
Improve
Like Article
Like
Save
Share
Report

Given an array arr[] and an integer k, we can cut this array at k different positions where k[] stores the positions of all the cuts required. The task is to print maximum sum among all the cuts after every cut made. Every cut is of the form of an integer x where x denotes a cut between arr[x] and arr[x + 1]

Examples:

Input: arr[] = {4, 5, 6, 7, 8}, k[] = {0, 2, 3, 1} 
Output: 26 15 11 8 
First cut -> {4} and {5, 6, 7, 8}. Maximum possible sum is 5 + 6 + 7 + 8 = 26 
Second cut -> {4}, {5, 6} and {7, 8}. Maximum sum = 15 
Third cut -> {4}, {5, 6}, {7} and {8}. Maximum sum = 11 
Fourth cut -> {4}, {5}, {6}, {7} and {8}. Maximum sum = 8 
Input: arr[] = {1, 2, 3}, k[] = {1} 
Output: 3

Naive approach: Store the resulting pieces of the array in an ArrayList and after every cut compute linearly the maximum possible sum. But this method would require O(n*k) time to answer all the queries. Efficient approach: We can represent each resulting piece of the array as a Piece object with data members start (start index of this piece), end (end index of this piece) and value (sum value of this piece). We can store these pieces in a TreeSet and sort them by their sum values. Therefore, after every cut we can get the Piece with largest sum value in O(log(n)).

  • We have to make a prefix sum array of the array values to get the sum between two indices in constant time.
  • We have to maintain another TreeSet with start indexes of all the current pieces so that we can find the exact piece to cut. For example, for a single piece:
    1. {1, 8} -> start = 1, end = 2, value = 9 and {6, 3, 9} -> start = 3, end = 5, value = 18.
    2. In order to cut index 4, we need to cut the second piece into two pieces as {6, 3} ans {9}. So we get the start index of which piece to cut from this TreeSet.

Below is the implementation of the above approach: 

Java




// Java implementation of the approach
import java.io.IOException;
import java.io.InputStream;
import java.util.*;
   
// Comparator to sort the Pieces
// based on their sum values
class MyComp implements Comparator<Piece> {
    public int compare(Piece p1, Piece p2)
    {
        if (p2.val != p1.val)
            return p2.val - p1.val;
        if (p1.start != p2.start)
            return p2.start - p1.start;
        return 0;
    }
}
class Piece {
    int start;
    int end;
    int val;
   
    // Constructor to initialize each Piece
    Piece(int s, int e, int v)
    {
        start = s;
        end = e;
        val = v;
    }
}
   
class GFG {
   
    // Function to perform the given queries on the array
    static void solve(int n, int k, int cuts[], int A[])
    {
   
        // Prefix sum array
        int sum[] = new int[n];
        sum[0] = A[0];
        for (int i = 1; i < n; i++)
            sum[i] = sum[i - 1] + A[i];
   
        // TreeSet storing all the starts
        TreeSet<Integer> t = new TreeSet<>();
   
        // TreeSet storing the actual pieces
        TreeSet<Piece> pq = new TreeSet<>(new MyComp());
        Piece temp[] = new Piece[n];
        temp[0] = new Piece(0, n - 1, sum[n - 1]);
   
        // Added the whole array or Piece of array
        // as there is no cuts yet
        pq.add(temp[0]);
        t.add(0);
   
        for (int i = 0; i < k; i++) {
   
            // curr is the piece to be cut
            int curr = t.floor(cuts[i]);
            pq.remove(temp[curr]);
            int end = temp[curr].end;
   
            // When a piece with start = s and end = e
            // is cut at index i, two pieces are created with
            // start = s, end = i and start = i + 1 and end = e
            // We remove the previous piece and add
            // this one to the TreeSet
            temp[curr]
                = new Piece(curr, cuts[i],
                            sum[cuts[i]]
                                - (curr == 0 ? 0 : sum[curr - 1]));
            pq.add(temp[curr]);
   
            temp[cuts[i] + 1]
                = new Piece(cuts[i] + 1,
                            end, sum[end] - sum[cuts[i]]);
            pq.add(temp[cuts[i] + 1]);
   
            t.add(curr);
            t.add(cuts[i] + 1);
   
            System.out.println(pq.first().val);
        }
    }
   
    // Driver code
    public static void main(String[] args)
    {
   
        int A[] = { 4, 5, 6, 7, 8 };
        int n = A.length;
        int cuts[] = { 0, 2, 3, 1 };
        int k = cuts.length;
   
        solve(n, k, cuts, A);
    }
}


Python3




# Python implementation of the approach
from functools import cmp_to_key
 
# Comparator to sort the Pieces
# based on their sum values
def my_comp(p1, p2):
    if p2.val != p1.val:
        return p2.val - p1.val
    if p1.start != p2.start:
        return p2.start - p1.start
    return 0
 
class Piece:
    def __init__(self, s, e, v):
        self.start = s
        self.end = e
        self.val = v
 
# Function to perform the given queries on the array
def solve(n, k, cuts, A):
    # Prefix sum array
    sum = [0] * n
    sum[0] = A[0]
    for i in range(1, n):
        sum[i] = sum[i - 1] + A[i]
 
    # TreeSet storing all the starts
    t = set()
 
    # TreeSet storing the actual pieces
    pq = []
    temp = [None] * n
    temp[0] = Piece(0, n - 1, sum[n - 1])
 
    # Added the whole array or Piece of array
    # as there is no cuts yet
    pq.append(temp[0])
    t.add(0)
 
    for i in range(k):
        # curr is the piece to be cut
        curr = max(t, key=lambda x: x if x <= cuts[i] else float('-inf'))
        pq.remove(temp[curr])
        end = temp[curr].end
 
        # When a piece with start = s and end = e
        # is cut at index i, two pieces are created with
        # start = s, end = i and start = i + 1 and end = e
        # We remove the previous piece and add
        # this one to the TreeSet
        temp[curr] = Piece(curr, cuts[i], sum[cuts[i]] - (0 if curr == 0 else sum[curr - 1]))
        pq.append(temp[curr])
 
        temp[cuts[i] + 1] = Piece(cuts[i] + 1, end, sum[end] - sum[cuts[i]])
        pq.append(temp[cuts[i] + 1])
 
        t.add(curr)
        t.add(cuts[i] + 1)
 
        pq.sort(key=cmp_to_key(my_comp))
        print(pq[0].val)
 
# Driver code
if __name__ == '__main__':
    A = [4, 5, 6, 7, 8]
    n = len(A)
    cuts = [0, 2, 3, 1]
    k = len(cuts)
    solve(n, k, cuts, A)


C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
// Comparator to sort the Pieces
// based on their sum values
class MyComp : IComparer<Piece> {
  public int Compare(Piece p1, Piece p2)
  {
    if (p2.val != p1.val)
      return p2.val - p1.val;
    if (p1.start != p2.start)
      return p2.start - p1.start;
    return 0;
  }
}
class Piece {
  public int start;
  public int end;
  public int val;
 
  // Constructor to initialize each Piece
  public Piece(int s, int e, int v)
  {
    start = s;
    end = e;
    val = v;
  }
}
 
class GFG {
 
  // Function to perform the given queries on the array
  static void solve(int n, int k, int[] cuts, int[] A)
  {
 
    // Prefix sum array
    int[] sum = new int[n];
    sum[0] = A[0];
    for (int i = 1; i < n; i++)
      sum[i] = sum[i - 1] + A[i];
 
    // TreeSet storing all the starts
    SortedSet<int> t = new SortedSet<int>();
 
    // TreeSet storing the actual pieces
    SortedSet<Piece> pq
      = new SortedSet<Piece>(new MyComp());
    Piece[] temp = new Piece[n];
    temp[0] = new Piece(0, n - 1, sum[n - 1]);
 
    // Added the whole array or Piece of array
    // as there is no cuts yet
    pq.Add(temp[0]);
    t.Add(0);
 
    for (int i = 0; i < k; i++) {
 
      // curr is the piece to be cut
      int curr
        = t.GetViewBetween(int.MinValue, cuts[i])
        .Max;
      pq.Remove(temp[curr]);
      int end = temp[curr].end;
 
      // When a piece with start = s and end = e
      // is cut at index i, two pieces are created
      // with start = s, end = i and start = i + 1 and
      // end = e We remove the previous piece and add
      // this one to the TreeSet
      temp[curr] = new Piece(
        curr, cuts[i],
        sum[cuts[i]]
        - (curr == 0 ? 0 : sum[curr - 1]));
      pq.Add(temp[curr]);
 
      temp[cuts[i] + 1] = new Piece(
        cuts[i] + 1, end, sum[end] - sum[cuts[i]]);
      pq.Add(temp[cuts[i] + 1]);
 
      t.Add(curr);
      t.Add(cuts[i] + 1);
 
      Console.WriteLine(pq.Min.val);
    }
  }
 
  // Driver code
  public static void Main(string[] args)
  {
 
    int[] A = { 4, 5, 6, 7, 8 };
    int n = A.Length;
    int[] cuts = { 0, 2, 3, 1 };
    int k = cuts.Length;
 
    // Function call
    solve(n, k, cuts, A);
  }
}
 
// This code is contributed by phasing17.


Javascript




// javascript implementation of the approach
 
 
 // Comparator to sort the Pieces
 // based on their sum values
class Piece{
     
    constructor(s, e, v){
        this.start = s;
        this.end= e;
        this.val = v;
    }
}
 
function upper_bound(t, x){
     
    let a = Array.from(t);
    a.sort();
    let l = 0;
    let h = a.length - 1;
     
    while(l <= h){
        let m = Math.floor((l+h)/2);
         
        if(a[m] < x){
            l = m + 1;
        }
        else if(a[m] == x) return a[m];
        else h = m - 1;
    }
     
    l--;
    return a[l];
}
 
// Function to perform the given queries on the array
function solve(n, k, cuts, A){
    // Prefix sum array
    let sum = new Array(n).fill(0);
    sum[0] = A[0];
    for(let i = 1; i < n; i++){
        sum[i] = sum[i-1] + A[i];
    }
 
 
    // TreeSet storing all the starts
    let t = new Set();
 
    // TreeSet storing the actual pieces
    let pq = [];
    let temp = new Array(n).fill(null);
    temp[0] = new Piece(0, n - 1, sum[n - 1]);
 
    // Added the whole array or Piece of array
    // as there is no cuts yet
    pq.push(temp[0]);
    t.add(0);
 
    for(let i = 0; i < k; i++){
        // curr is the piece to be cut
        // console.log(cuts + "  " + i);
        let curr = upper_bound(t, cuts[i]);
        pq.splice(pq.indexOf(temp[curr]), 1);
        let end = temp[curr].end;
 
        // When a piece with start = s and end = e
        // is cut at index i, two pieces are created with
        // start = s, end = i and start = i + 1 and end = e
        // We remove the previous piece and add
        // this one to the TreeSet
        temp[curr] = new Piece(curr, cuts[i], sum[cuts[i]] - ((curr == 0)? 0: sum[curr - 1]));
        pq.push(temp[curr]);
 
        temp[cuts[i] + 1] = new Piece(cuts[i] + 1, end, sum[end] - sum[cuts[i]]);
        pq.push(temp[cuts[i] + 1]);
 
        t.add(curr);
        t.add(cuts[i] + 1);
 
        pq.sort(function(p1, p2){
            if(p2.val != p1.val)
                return p2.val - p1.val;
            if(p1.start != p2.start)
                return p2.start - p1.start;
            return 0;
        })
        console.log(pq[0].val);
    }
}
 
// #Driver code
let A = [4, 5, 6, 7, 8];
let n = A.length
let cuts = [0, 2, 3, 1];
let k = cuts.length;
solve(n, k, cuts, A);
 
// The code is contributed by Nidhi goel.


Output:

26
15
11
8

Time Complexity O(n + k Log n)

Space Complexity: O(n + k)



Last Updated : 08 May, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads