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Find den maksimale værdi af abs(i - j) * min(arr[i], arr[j]) i en matrix arr[]

Givet en række af n forskellige elementer. Find maksimum for produktet af Minimum af to tal i matrixen og absolut forskel på deres positioner, dvs. find maksimumværdien af ​​abs(i - j) * min(arr[i] arr[j]), hvor i og j varierer fra 0 til n-1. 

brug af operativsystem

Eksempler:  

Input : arr[] = {3 2 1 4} Output: 9 // arr[0] = 3 and arr[3] = 4 minimum of them is 3 and // absolute difference between their position is // abs(0-3) = 3. So product is 3*3 = 9 Input : arr[] = {8 1 9 4} Output: 16 // arr[0] = 8 and arr[2] = 9 minimum of them is 8 and // absolute difference between their position is // abs(0-2) = 2. So product is 8*2 = 16
Recommended Practice Find maksimal værdi Prøv det!

EN enkel løsning for dette problem er at tage hvert element et efter et og sammenligne dette element med elementerne til højre for det. Beregn derefter produktet af minimum af dem og absolut forskel mellem deres indekser og maksimer resultatet. Tidskompleksiteten for denne tilgang er O(n^2).



An effektiv løsning at løse problemet i lineær tidskompleksitet. Vi tager to iteratorer Venstre=0 og Højre=n-1 sammenlign elementerne arr[Venstre] og arr[højre].  

left = 0 right = n-1 maxProduct = -INF While (left < right) If arr[Left] < arr[right] currProduct = arr[Left]*(right-Left) Left++ . If arr[right] < arr[Left] currProduct = arr[Right]*(Right-Left) Right-- . maxProduct = max(maxProduct currProduct)

Nedenfor er implementeringen af ​​ovenstående idé. 

C++ 
// C++ implementation of code #include   using namespace std; // Function to calculate maximum value of  // abs(i - j) * min(arr[i] arr[j]) in arr[] int Maximum_Product(int arr[] int n) {  int maxProduct = INT_MIN; // Initialize result  int currProduct; // product of current pair  // loop until they meet with each other  int Left = 0 right = n-1;  while (Left < right)  {  if (arr[Left] < arr[right])  {  currProduct = arr[Left]*(right-Left);  Left++;  }  else // arr[right] is smaller  {  currProduct = arr[right]*(right-Left);  right--;  }  // maximizing the product  maxProduct = max(maxProduct currProduct);  }  return maxProduct; } // Driver program to test the case int main() {  int arr[] = {8 1 9 4};  int n = sizeof(arr)/sizeof(arr[0]);  cout << Maximum_Product(arrn);  return 0; }
Java 
// Java implementation of code import java.util.*; class GFG {    // Function to calculate maximum value of  // abs(i - j) * min(arr[i] arr[j]) in arr[]  static int Maximum_Product(int arr[] int n) {    // Initialize result  int maxProduct = Integer.MIN_VALUE;     // product of current pair  int currProduct;   // loop until they meet with each other  int Left = 0 right = n - 1;  while (Left < right) {  if (arr[Left] < arr[right]) {  currProduct = arr[Left] * (right - Left);  Left++;  }     // arr[right] is smaller  else   {  currProduct = arr[right] * (right - Left);  right--;  }  // maximizing the product  maxProduct = Math.max(maxProduct currProduct);  }  return maxProduct; } // Driver code public static void main(String[] args)  {  int arr[] = {8 1 9 4};  int n = arr.length;  System.out.print(Maximum_Product(arr n)); } } // This code is contributed by Anant Agarwal.
Python3 
# Python implementation of code # Function to calculate # maximum value of  # abs(i - j) * min(arr[i] # arr[j]) in arr[] def Maximum_Product(arrn): # Initialize result maxProduct = -2147483648 # product of current pair currProduct=0 # loop until they meet with each other Left = 0 right = n-1 while (Left < right): if (arr[Left] < arr[right]): currProduct = arr[Left]*(right-Left) Left+=1 else: # arr[right] is smaller currProduct = arr[right]*(right-Left) right-=1 # maximizing the product maxProduct = max(maxProduct currProduct) return maxProduct # Driver code arr = [8 1 9 4] n = len(arr) print(Maximum_Product(arrn)) # This code is contributed # by Anant Agarwal.
C# 
// C# implementation of code using System; class GFG {   // Function to calculate maximum // value of abs(i - j) * min(arr[i] // arr[j]) in arr[] static int Maximum_Product(int []arr  int n) {    // Initialize result  int maxProduct = int.MinValue;     // product of current pair  int currProduct;   // loop until they meet   // with each other  int Left = 0 right = n - 1;  while (Left < right) {  if (arr[Left] < arr[right])  {  currProduct = arr[Left] *   (right - Left);  Left++;  }     // arr[right] is smaller  else  {  currProduct = arr[right] *  (right - Left);  right--;  }  // maximizing the product  maxProduct = Math.Max(maxProduct   currProduct);  }  return maxProduct; } // Driver code public static void Main()  {  int []arr = {8 1 9 4};  int n = arr.Length;  Console.Write(Maximum_Product(arr n)); } } // This code is contributed by nitin mittal.
PHP 
 // PHP implementation of code // Function to calculate  // maximum value of  // abs(i - j) * min(arr[i]  // arr[j]) in arr[] function Maximum_Product($arr $n) { $INT_MIN = 0; // Initialize result $maxProduct = $INT_MIN; // product of current pair $currProduct; // loop until they meet // with each other $Left = 0; $right = $n - 1; while ($Left < $right) { if ($arr[$Left] < $arr[$right]) { $currProduct = $arr[$Left] * ($right - $Left); $Left++; } // arr[right] is smaller else { $currProduct = $arr[$right] * ($right - $Left); $right--; } // maximizing the product $maxProduct = max($maxProduct $currProduct); } return $maxProduct; } // Driver Code $arr = array(8 1 9 4); $n = sizeof($arr) / sizeof($arr[0]); echo Maximum_Product($arr $n); // This code is contributed // by nitin mittal.  ?>
JavaScript 
<script> // Javascript implementation of code // Function to calculate // maximum value of // abs(i - j) * min(arr[i] // arr[j]) in arr[] function Maximum_Product(arr n) {  let INT_MIN = 0;  // Initialize result  let maxProduct = INT_MIN;  // Product of current pair  let currProduct;  // Loop until they meet  // with each other  let Left = 0 right = n - 1;  while (Left < right)   {  if (arr[Left] < arr[right])  {  currProduct = arr[Left] *  (right - Left);  Left++;  }  // arr[right] is smaller  else   {  currProduct = arr[right] *  (right - Left);  right--;  }  // Maximizing the product  maxProduct = Math.max(maxProduct  currProduct);  }  return maxProduct; } // Driver Code let arr = new Array(8 1 9 4); let n = arr.length; document.write(Maximum_Product(arr n)); // This code is contributed by Saurabh Jaiswal </script>

Produktion 
16

Tidskompleksitet: O(N log N) her er N længden af ​​Array.

Rumkompleksitet: O(1) da der ikke bruges ekstra plads.

Hvordan virker dette?  
Det vigtige for at vise, at vi ikke går glip af noget potentielt par i ovenstående lineære algoritme, dvs. vi er nødt til at vise, at at gøre venstre++ eller højre-- ikke fører til et tilfælde, hvor vi ville have fået en højere værdi af maxProduct.

Bemærk, at vi altid gange med (højre - venstre). 

  1. Hvis arr[venstre]< arr[right] then smaller values of højre for nuværende venstre er ubrugelige, da de ikke kan producere højere værdi af maxProduct (fordi vi multiplicerer med arr[venstre] med (højre - venstre)). Hvad hvis arr[venstre] var større end nogen af ​​elementerne på dens venstre side. I så fald skal der være fundet et bedre par for det element med nuværende ret. Derfor kan vi roligt øge venstre uden at gå glip af noget bedre par med strøm til venstre.
  2. Lignende argumenter er anvendelige, når arr[right]< arr[left].