Find the value of max(f(x)) – min(f(x)) for a given F(x)

Given two sequences of N integers consisting of both positive and negative integers, inclusive of 0 (sequence A and sequence B). Then there will be Q queries. In each, you will be given two integers l and r (r>=l). Let’s define a function: 

The task is to print the value of Max(f(x, y)) – Min(f(x, y)) for each query.

Examples:

Input: 
N = 5, Q = 2
A[] = 0 7 3 4 5
B[] = 0 3 1 2 3
l = 1, r = 1
l = 1, r = 3
Output: 
0
917448

Input: 
N = 5, Q = 2
A[] = 0 -8 3 4 -9
B[] = 0 -3 -5 2 3
l = 1, r = 1
l = 1, r = 3
Output:
0
851916 

Approach: If we clearly see, the function only depends on the value of x and y which can be easily calculated by seeing the nature of the summation of the individual arrays. First, we precalculate the prefix sum of both the arrays, and then we just see the nature of the summations. And always the value of x and y will either be 32766 or -32766 as we are taking the modulo of them.

Below is the implementation of the above approach:

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define MAX 200006
#define CONS 32766

// Function to calculate the value
void calc(ll a[], ll b[], ll lr[], ll q, ll n)
{
    ll M, m, i, j, k, l, r, suma, sumb, cc;
    cc = 0;

    // forming the prefix sum arrays
    for (i = 0; i < n - 1; ++i) {
        a[i + 1] += a[i];
        b[i + 1] += b[i];
    }

    while (q--) {
        // Taking the query
        l = lr[cc++];
        r = lr[cc++];
        l -= 2;
        r -= 1;

        // finding the sum in the range l to r in array a
        suma = a[r];

        // finding the sum in the range l to r in array b
        sumb = b[r];
        if (l >= 0) {
            suma -= a[l];
            sumb -= b[l];
        }

        // Finding the max value of the function
        M = max(CONS * suma + CONS * sumb,
                -CONS * suma - CONS * sumb);
        M = max(M, max(CONS * suma - CONS * sumb,
                       -CONS * suma + CONS * sumb));

        // Finding the min value of the function
        m = min(CONS * suma + CONS * sumb,
                -CONS * suma - CONS * sumb);
        m = min(m, min(CONS * suma - CONS * sumb,
                       -CONS * suma + CONS * sumb));

        cout << (M - m) << "\n";
    }
}

// Driver code
int main()
{
    ll n = 5, q = 2;
    ll a[5] = { 0, 7, 3, 4, 5 };
    ll b[5] = { 0, 3, 1, 2, 3 };

    ll lr[q * 2];
    lr[0] = 1;
    lr[1] = 1;
    lr[2] = 1;
    lr[3] = 3;

    calc(a, b, lr, q, n);

    return 0;
}

Java

// Java implementation of above approach

import java.util.*;

class GFG
{

        static final int MAX=200006;
        static final int CONS=32766;

        // Function to calculate the value
        static void calc(int a[], int b[], int lr[], int q, int n)
        {
            int M, m, i, j, k, l, r, suma, sumb, cc;
            cc = 0;

            // forming the prefix sum arrays
            for (i = 0; i < n - 1; ++i)
            {
                a[i + 1] += a[i];
                b[i + 1] += b[i];
            }

            while (q!=0)
            {
                // Taking the query
                l = lr[cc++];
                r = lr[cc++];
                l -= 2;
                r -= 1;

                // finding the sum in the range l to r in array a
                suma = a[r];

                // finding the sum in the range l to r in array b
                sumb = b[r];
                if (l >= 0) {
                    suma -= a[l];
                    sumb -= b[l];
                }

                // Finding the max value of the function
                M = Math.max(CONS * suma + CONS * sumb,
                        -CONS * suma - CONS * sumb);
                M = Math.max(M, Math.max(CONS * suma - CONS * sumb,
                            -CONS * suma + CONS * sumb));

                // Finding the min value of the function
                m = Math.min(CONS * suma + CONS * sumb,
                        -CONS * suma - CONS * sumb);
                m = Math.min(m, Math.min(CONS * suma - CONS * sumb,
                            -CONS * suma + CONS * sumb));

                System.out.println((M - m) );
                q--;
            }
        }

        // Driver code
        public static void  main(String [] args)
        {
            int n = 5, q = 2;
            int []a = { 0, 7, 3, 4, 5 };
            int []b = { 0, 3, 1, 2, 3 };

            int []lr=new int[q * 2];
            lr[0] = 1;
            lr[1] = 1;
            lr[2] = 1;
            lr[3] = 3;

            calc(a, b, lr, q, n);


        }
// This code is contributed by ihritik

}

Python3

# Python 3 implementation of
# above approach

MAX = 200006
CONS = 32766

# Function to calculate the value
def calc(a, b, lr, q, n):

    cc = 0

    # forming the prefix sum arrays
    for i in range(n - 1) :
        a[i + 1] += a[i]
        b[i + 1] += b[i]

    while (q > 0) :
        # Taking the query
        l = lr[cc]
        cc +=1
        r = lr[cc]
        cc += 1
        l -= 2
        r -= 1

        # finding the sum in the range l
        # to r in array a
        suma = a[r]

        # finding the sum in the range
        # l to r in array b
        sumb = b[r]
        if (l >= 0) :
            suma -= a[l]
            sumb -= b[l]

        # Finding the max value of the function
        M = max(CONS * suma + CONS * sumb,
                -CONS * suma - CONS * sumb)
        M = max(M, max(CONS * suma - CONS * sumb,
                    -CONS * suma + CONS * sumb))

        # Finding the min value of the function
        m = min(CONS * suma + CONS * sumb,
                -CONS * suma - CONS * sumb)
        m = min(m, min(CONS * suma - CONS * sumb,
                    -CONS * suma + CONS * sumb))

        print(M - m)

        q -= 1

# Driver code
if __name__ == "__main__":

    n = 5
    q = 2
    a = [ 0, 7, 3, 4, 5 ]
    b = [ 0, 3, 1, 2, 3 ]

    lr = [0]*(q * 2)
    lr[0] = 1
    lr[1] = 1
    lr[2] = 1
    lr[3] = 3

    calc(a, b, lr, q, n)

# This code is contributed by
# ChitraNayal

C#

// C# implementation of above approach
using System;

class GFG
{

// static int MAX=200006;
static int CONS = 32766;

// Function to calculate the value
static void calc(int []a, int []b,
                 int []lr, int q, int n)
{
    int M, m, i, l, r, suma, sumb, cc;
    cc = 0;

    // forming the prefix sum arrays
    for (i = 0; i < n - 1; ++i)
    {
        a[i + 1] += a[i];
        b[i + 1] += b[i];
    }

    while (q != 0)
    {
        // Taking the query
        l = lr[cc++];
        r = lr[cc++];
        l -= 2;
        r -= 1;

        // finding the sum in the
        // range l to r in array a
        suma = a[r];

        // finding the sum in the
        // range l to r in array b
        sumb = b[r];
        if (l >= 0)
        {
            suma -= a[l];
            sumb -= b[l];
        }

        // Finding the max value of the function
        M = Math.Max(CONS * suma + CONS * sumb,
                    -CONS * suma - CONS * sumb);
        M = Math.Max(M, Math.Max(CONS * suma - CONS * sumb,
                                -CONS * suma + CONS * sumb));

        // Finding the min value of the function
        m = Math.Min(CONS * suma + CONS * sumb,
                    -CONS * suma - CONS * sumb);
        m = Math.Min(m, Math.Min(CONS * suma - CONS * sumb,
                                -CONS * suma + CONS * sumb));

        Console.WriteLine((M - m));
        q--;
    }
}

// Driver code
public static void Main()
{
    int n = 5, q = 2;
    int []a = { 0, 7, 3, 4, 5 };
    int []b = { 0, 3, 1, 2, 3 };

    int []lr = new int[q * 2];
    lr[0] = 1;
    lr[1] = 1;
    lr[2] = 1;
    lr[3] = 3;

    calc(a, b, lr, q, n);
}
}

// This code is contributed by anuj_67

PHP

<?php
// PHP implementation of above approach
$MAX = 200006;
$CONS = 32766;

// Function to calculate the value
function calc($a, $b, $lr, $q, $n)
{
    global $MAX ;
    global $CONS;
    $M; $m; $i; $j; $k;
    $l; $r; $suma; $sumb; $cc;
    $cc = 0;

    // forming the prefix sum arrays
    for ($i = 0; $i < $n - 1; ++$i)
    {
        $a[$i + 1] += $a[$i];
        $b[$i + 1] += $b[$i];
    }

    while ($q--)
    {
        // Taking the query
        $l = $lr[$cc++];
        $r = $lr[$cc++];
        $l -= 2;
        $r -= 1;

        // finding the sum in the range
        // l to r in array a
        $suma = $a[$r];

        // finding the sum in the range
        // l to r in array b
        $sumb = $b[$r];
        if ($l >= 0)
        {
            $suma -= $a[$l];
            $sumb -= $b[$l];
        }

        // Finding the max value of the function
        $M = max($CONS * $suma + $CONS * $sumb,
                -$CONS * $suma - $CONS * $sumb);
        $M = max($M, max($CONS * $suma - $CONS * $sumb,
                        -$CONS * $suma + $CONS * $sumb));

        // Finding the min value of the function
        $m = min($CONS * $suma + $CONS * $sumb,
                -$CONS * $suma - $CONS * $sumb);
        $m = min($m, min($CONS * $suma - $CONS * $sumb,
                        -$CONS * $suma + $CONS * $sumb));

        echo ($M - $m) , "\n";
    }
}

// Driver code
$n = 5; $q = 2;
$a = array(0, 7, 3, 4, 5 );
$b = array( 0, 3, 1, 2, 3 );
$lr[0] = 1;
$lr[1] = 1;
$lr[2] = 1;
$lr[3] = 3;

calc($a, $b, $lr, $q, $n);

// This code is contributed by anuj_67
?>

Javascript

<script>
// Javascript implementation of above approach

    let MAX=200006;
    let CONS=32766;
    // Function to calculate the value
    function calc(a,b,lr,q,n)
    {
            let M, m, i, j, k, l, r, suma, sumb, cc;
            cc = 0;

            // forming the prefix sum arrays
            for (i = 0; i < n - 1; ++i)
            {
                a[i + 1] += a[i];
                b[i + 1] += b[i];
            }

            while (q!=0)
            {
                // Taking the query
                l = lr[cc++];
                r = lr[cc++];
                l -= 2;
                r -= 1;

                // finding the sum in the range l to r in array a
                suma = a[r];

                // finding the sum in the range l to r in array b
                sumb = b[r];
                if (l >= 0) {
                    suma -= a[l];
                    sumb -= b[l];
                }

                // Finding the max value of the function
                M = Math.max(CONS * suma + CONS * sumb,
                        -CONS * suma - CONS * sumb);
                M = Math.max(M, Math.max(CONS * suma - CONS * sumb,
                            -CONS * suma + CONS * sumb));

                // Finding the min value of the function
                m = Math.min(CONS * suma + CONS * sumb,
                        -CONS * suma - CONS * sumb);
                m = Math.min(m, Math.min(CONS * suma - CONS * sumb,
                            -CONS * suma + CONS * sumb));

                document.write((M - m) +"<br>");
                q--;
            }
    }

    // Driver code
    let n = 5, q = 2;
    let a=[ 0, 7, 3, 4, 5 ];
    let b=[0, 3, 1, 2, 3];
    let lr=new Array(q * 2);
    lr[0] = 1;
    lr[1] = 1;
    lr[2] = 1;
    lr[3] = 3;

    calc(a, b, lr, q, n);

// This code is contributed by avanitrachhadiya2155
</script>

Output

0
917448

Complexity analysis:

• Time Complexity: O(n) // only one traversal of the array is needed.\
• Auxiliary Complexity: O(q*2)