Nájdite maximálnu hodnotu abs(i - j) * min(arr[i], arr[j]) v poli arr[]
Dané pole n rôznych prvkov. Nájdite maximum súčinu minima dvoch čísel v poli a absolútny rozdiel ich pozícií, t. j. nájdite maximálnu hodnotu abs(i - j) * min(arr[i] arr[j]), kde i a j sa menia od 0 do n-1.
Príklady:
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 Nájdite maximálnu hodnotu Skúste to! A jednoduché riešenie pretože tento problém je vziať každý prvok jeden po druhom a porovnať tento prvok s prvkami napravo od neho. Potom vypočítajte súčin minima z nich a absolútneho rozdielu medzi ich indexmi a maximalizujte výsledok. Časová zložitosť pre tento prístup je O(n^2).
An efektívne riešenie riešiť problém v lineárnej časovej zložitosti. Vezmeme dva iterátory Vľavo = 0 a vpravo=n-1 porovnajte prvky arr[Left] a arr[right].
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)
Nižšie je uvedená implementácia vyššie uvedenej myšlienky.
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 ( arr n ); 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 ( arr n ): # 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 ( arr n )) # 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>
Výstup
16
Časová zložitosť: O(N log N) tu N je dĺžka poľa.
Priestorová zložitosť: O(1) pretože sa nepoužíva žiadny ďalší priestor.
Ako to funguje?
Dôležité je ukázať, že nám vo vyššie uvedenom lineárnom algoritme nechýba žiadny potenciálny pár, t.j. musíme ukázať, že robenie vľavo ++ alebo vpravo nevedie k prípadu, kedy by sme dostali vyššiu hodnotu maxProduct.
Upozorňujeme, že vždy násobíme pomocou (vpravo - vľavo).
- Ak arr[vľavo] < arr[right] then smaller values of správne pre prúd vľavo sú zbytočné, pretože nemôžu produkovať vyššiu hodnotu maxProduct (pretože násobíme s arr[left] s (right - left)). Čo ak arr[left] bol väčší ako ktorýkoľvek z prvkov na jeho ľavej strane. V takom prípade sa musí nájsť lepší pár pre tento prvok s aktuálnym právom. Preto môžeme bezpečne zvýšiť vľavo bez toho, aby sme vynechali lepší pár so ľavým prúdom.
- Podobné argumenty sú použiteľné, keď arr[vpravo] < arr[left].
