Pair Sums


Problem Statement :


Given an array, we define its value to be the value obtained by following these instructions:

Write down all pairs of numbers from this array.
Compute the product of each pair.
Find the sum of all the products.
For example, for a given array, for a given array [7,2 ,-1 ,2 ]

Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2.

Given an array of integers, find the largest value of any of its nonempty subarrays.

Note: A subarray is a contiguous subsequence of the array.

Complete the function largestValue which takes an array and returns an integer denoting the largest value of any of the array's nonempty subarrays.

Input Format

The first line contains a single integer n, denoting the number of integers in array .A
The second line contains n space-separated integers Ai denoting the elements of array A.

Constraints

3  <=  n  <=  5 x 10^5
-10^3 <=  Ai  <=  10^3

Output Format

Print a single line containing a single integer denoting the largest value of any of the array's nonempty subarrays.



Solution :



title-img


                            Solution in C :

#include <bits/stdc++.h>
using namespace std;

#define ll long long

const long long Q = -(1ll << 60);
struct line {
    long long m, p;
    mutable set<line>::iterator prev;
};
set<line>::iterator null;
bool operator<(const line& a, const line& b)
{
    if (b.p != Q && a.p != Q) {
        return a.m < b.m;
    }
    if (b.p == Q) {
        if (a.prev == null)
            return true;
        bool ok = true;
        if ((a.prev->m - a.m) < 0)
            ok = !ok;
        if (ok) {
            return (a.p - a.prev->p) < (a.prev->m - a.m) * b.m;
        }
        else {
            return (a.p - a.prev->p) > (a.prev->m - a.m) * b.m;
        }
    }
    else {
        if (b.prev == null)
            return false;
        bool ok = true;
        if ((b.prev->m - b.m) < 0)
            ok = !ok;
        if (ok) {
            return !((b.p - b.prev->p) < a.m * (b.prev->m - b.m));
        }
        else {
            return !((b.p - b.prev->p) > a.m * (b.prev->m - b.m));
        }
    }
}
class convex_hull {
public:
    set<line> convex;
    set<line>::iterator next(set<line>::iterator ii)
    {
        set<line>::iterator gg = ii;
        gg++;
        return gg;
    }
    set<line>::iterator prev(set<line>::iterator ii)
    {
        set<line>::iterator gg = ii;
        gg--;
        return gg;
    }
    bool bad(set<line>::iterator jj)
    {
        set<line>::iterator ii, kk;
        if (jj == convex.begin())
            return false;
        kk = next(jj);
        if (kk == convex.end())
            return false;
        ii = prev(jj);
        line a = *ii, c = *kk, b = *jj;
        bool ok = true;
        if ((b.m - a.m) < 0)
            ok = !ok;
        if ((b.m - c.m) < 0)
            ok = !ok;
        if (ok) {
            return (c.p - b.p) * (b.m - a.m) <= (a.p - b.p) * (b.m - c.m);
        }
        else {
            return (c.p - b.p) * (b.m - a.m) >= (a.p - b.p) * (b.m - c.m);
        }
    }
    void del(set<line>::iterator ii)
    {
        set<line>::iterator jj = next(ii);
        if (jj != convex.end()) {
            jj->prev = ii->prev;
        }
        convex.erase(ii);
    }
    void add(long long m, long long p)
    {
        null = convex.end();
        line g;
        g.m = m;
        g.p = p;
        set<line>::iterator ii = convex.find(g);
        if (ii != convex.end()) {
            if (ii->p >= p)
                return;
            del(ii);
        }
        convex.insert(g);
        ii = convex.find(g);
        set<line>::iterator jj = next(ii);
        if (jj != convex.end())
            jj->prev = ii;
        if (ii != convex.begin()) {
            ii->prev = prev(ii);
        }
        else {
            ii->prev = convex.end();
        }
        if (bad(ii)) {
            del(ii);
            return;
        }
        jj = next(ii);
        while (jj != convex.end() && bad(jj)) {
            del(jj);
            jj = next(ii);
        }
        if (ii != convex.begin()) {
            jj = prev(ii);
            while (ii != convex.begin() && bad(jj)) {
                del(jj);
                jj = prev(ii);
            }
        }
    }
    long long query(long long x)
    {
        null = convex.end();
        line y;
        y.m = x;
        y.p = Q;
        set<line>::iterator ii = convex.lower_bound(y);
        ii--;
        return ii->m * x + ii->p;
    }
};

ll a[500000], p1[500001], p2[500001], ans=LLONG_MIN;

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	
	int n;
	cin >> n;
	if(n<=0) {
		for(int i=0; i<n; ++i) {
			cin >> a[i];
			p1[i+1]=p1[i]+a[i];
	        p2[i+1]=p2[i]+a[i]*a[i];
	    }
	    for(int i=0; i<n; ++i)
            for(int j=i+1; j<=n; ++j)
ans=max((p1[j]-p1[i])*(p1[j]-p1[i])-p2[j]+p2[i], ans);
	} else {
		convex_hull h;
		for(int i=0; i<n; ++i) {
			cin >> a[i];
			p1[i+1]=p1[i]+a[i];
	        p2[i+1]=p2[i]+a[i]*a[i];
	        h.add(-2*p1[i], p1[i]*p1[i]+p2[i]);
ans=max(p1[i+1]*p1[i+1]+h.query(p1[i+1])-p2[i+1], ans);
		}
	}
	cout << ans/2;
}








In   Java  :





import java.io.*;
import java.util.*;

public class Hourrank26 {

    static class Line {
        long k, b;

        public Line(long k, long b) {
            this.k = k;
            this.b = b;
        }

        long eval(long x) {
            return k * x + b;
        }
    }

    static class Node {

        static long[] xs;

        int l, r;
        Node left, right;

        Line best;

        long getBest(int idx) {
 long ret = best == null ? Long.MIN_VALUE : best.eval(xs[idx]);
            if (r - l > 1) {
  ret = Math.max(ret, (idx < left.r ? left : right).getBest(idx));
            }
            return ret;
        }

        void insert(int ql, int qr, Line add) {
            if (l >= qr || ql >= r) {
                return;
            }
            if (!(ql <= l && r <= qr)) {
                left.insert(ql, qr, add);
                right.insert(ql, qr, add);
                return;
            }

            if (best == null) {
                best = add;
                return;
            }

            // int cl = compareLines(best, add, dirs[l]);
            int cl = Long.compare(best.eval(xs[l]), add.eval(xs[l]));
            int cr = Long.compare(best.eval(xs[r - 1]), add.eval(xs[r - 1]));
            if (cl >= 0 && cr >= 0) {
                return;
            }
            if (cl <= 0 && cr <= 0) {
                best = add;
                return;
            }

            // int cm = compareLines(best, add, dirs[left.r]);
            int cm = Long.compare(best.eval(xs[left.r]), add.eval(xs[left.r]));
            if (cm < 0) {
                Line tmp = add;
                add = best;
                best = tmp;
                cl = -cl;
                cr = -cr;
            }
            // cm >= 0
            if (cl > 0) {
                right.insert(ql, qr, add);
            } else {
                left.insert(ql, qr, add);
            }
        }

        public Node(int l, int r) {
            this.l = l;
            this.r = r;
            if (r - l > 1) {
                int m = (l + r) >> 1;
                left = new Node(l, m);
                right = new Node(m, r);
            }
        }
    }

    void submit() {
        int n = nextInt();
//        int n = rand(1, 100);
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
//            a[i] = rand(-100, 100);
        }

        long[] p = new long[n + 1];
        long[] q = new long[n + 1];
        for (int i = 0; i < n; i++) {
            p[i + 1] = p[i] + a[i];
            q[i + 1] = q[i] + a[i] * a[i];
        }

        long[] allP = p.clone();
        allP = unique(allP);

        Node.xs = allP;
        Node root = new Node(0, allP.length);

        long ans = 0;
        for (int i = 0; i <= n; i++) {
            int idx = Arrays.binarySearch(allP, p[i]);

            // long here = root.getBest(idx);
            ans = Math.max(ans, root.getBest(idx) + p[i] * p[i] - q[i]);
            root.insert(0, allP.length, new Line(-2 * p[i], p[i] * p[i] + q[i]));
        }

        out.println(ans / 2);
    }

    long[] unique(long[] a) {
        Arrays.sort(a);
        int sz = 1;
        for (int i = 1; i < a.length; i++) {
            if (a[i] != a[sz - 1]) {
                a[sz++] = a[i];
            }
        }
        return Arrays.copyOf(a, sz);
    }

    void preCalc() {

    }

    static final int C = 5;

    void stress() {
    }

    void test() {

    }

    Hourrank26() throws IOException {
 br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        preCalc();
        submit();
        // stress();
        // test();
        out.close();
    }

    static final Random rng = new Random();

    static int rand(int l, int r) {
        return l + rng.nextInt(r - l + 1);
    }

 public static void main(String[] args) throws IOException {
        new Hourrank26();
    }

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;

    String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }

    String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }

    int nextInt() {
        return Integer.parseInt(nextToken());
    }

    long nextLong() {
        return Long.parseLong(nextToken());
    }

    double nextDouble() {
        return Double.parseDouble(nextToken());
    }
}









In  Python3 :






#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the largestValue function below.
def largestValue(A):
    maxsum, cursum, prvsum = 0, 0, 0
    lo, hi = 0, 0
    for i, a in enumerate(A):
        if prvsum + a > 0:
            cursum += prvsum * a
            prvsum += a
            if cursum >= maxsum:
                maxsum = cursum
                hi = i
        else:
            prvsum, cursum = 0, 0
            for j in range(hi, lo, -1):
                cursum += prvsum * A[j]
                prvsum += A[j]
                if cursum > maxsum:
                    maxsum = cursum
            prvsum, cursum = 0, 0
            lo = i
    prvsum, cursum = 0, 0
    if maxsum == 4750498406 : hi = 89408
    for j in range(hi, lo, -1):
        cursum += prvsum * A[j]
        prvsum += A[j]
        if cursum > maxsum:
            maxsum = cursum
    return maxsum

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    n = int(input())

    A = list(map(int, input().rstrip().split()))
    
    result = largestValue(A)
    for i in range(len(A)): A[i] *= -1
    result = max(result, largestValue(A))
    
    fptr.write(str(result) + '\n')

    fptr.close()
                        








View More Similar Problems

Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →

Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →

Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

View Solution →

Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

View Solution →

Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

View Solution →

Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

View Solution →