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

• Difficulty Level : Basic
• Last Updated : 20 May, 2021

Given two sequences of N integers consisting of both positive and negative integers, inclusive 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 using namespace std;#define ll long long#define MAX 200006#define CONS 32766 // Function to calculate the valuevoid 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 codeint 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 = 200006CONS = 32766 # Function to calculate the valuedef 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 codeif __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 approachusing System; class GFG{ // static int MAX=200006;static int CONS = 32766; // Function to calculate the valuestatic 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 codepublic 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

 = 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

 
Output:
0
917448

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