# 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 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



## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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

 = 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  ?>

Output:

0
917448


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