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, 8} -> start = 1, end = 2, value = 9 and {6, 3, 9} -> start = 3, end = 5, value = 18.
- 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 approachimport java.io.IOException;import java.io.InputStream;import java.util.*; // Comparator to sort the Pieces// based on their sum valuesclass 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 approachfrom functools import cmp_to_key# Comparator to sort the Pieces# based on their sum valuesdef 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 0class Piece: def __init__(self, s, e, v): self.start = s self.end = e self.val = v# Function to perform the given queries on the arraydef 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 codeif __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 approachusing System;using System.Collections.Generic;// Comparator to sort the Pieces// based on their sum valuesclass 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 valuesclass 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 arrayfunction 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 codelet A = [4, 5, 6, 7, 8];let n = A.lengthlet 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)
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