#practiceLinkDiv { display: none !important; }Duota n teigiamų skirtingų sveikųjų skaičių masyvas. Problema yra rasti didžiausią gretimų didėjančių pogrupių sumą O (n) laiko sudėtingumu.
Pavyzdžiai:
Input : arr[] = {2 1 4 7 3 6}Recommended Practice Godus Fox Išbandykite!
Output : 12
Contiguous Increasing subarray {1 4 7} = 12
Input : arr[] = {38 7 8 10 12}
Output : 38
A paprastas sprendimas yra generuoti visus pogrupius ir apskaičiuokite jų sumas. Galiausiai grąžinkite pogrupį su maksimalia suma. Šio sprendimo laiko sudėtingumas yra O(n2).
java žemėlapiai
An efektyvus sprendimas yra pagrįsta tuo, kad visi elementai yra teigiami. Taigi mes atsižvelgiame į ilgiausiai didėjančias pogrupes ir palyginame jų sumas. Didėjantis posluoksnis negali sutapti, todėl mūsų laiko sudėtingumas tampa O (n).
Algoritmas:
Let arr be the array of size n
Let result be the required sum
int largestSum(arr n)
result = INT_MIN // Initialize result
i = 0
while i < n
// Find sum of longest increasing subarray
// starting with i
curr_sum = arr[i];
while i+1 < n && arr[i] < arr[i+1]
curr_sum += arr[i+1];
i++;
// If current sum is greater than current
// result.
if result < curr_sum
result = curr_sum;
i++;
return result
Žemiau pateikiamas aukščiau pateikto algoritmo įgyvendinimas.
java sąrašo mazgasC++
// C++ implementation of largest sum // contiguous increasing subarray #include
// Java implementation of largest sum // contiguous increasing subarray class GFG { // Returns sum of longest // increasing subarray. static int largestSum(int arr[] int n) { // Initialize result int result = -9999999; // Note that i is incremented // by inner loop also so overall // time complexity is O(n) for (int i = 0; i < n; i++) { // Find sum of longest // increasing subarray // starting from arr[i] int curr_sum = arr[i]; while (i + 1 < n && arr[i + 1] > arr[i]) { curr_sum += arr[i + 1]; i++; } // Update result if required if (curr_sum > result) result = curr_sum; } // required largest sum return result; } // Driver Code public static void main(String[] args) { int arr[] = { 1 1 4 7 3 6 }; int n = arr.length; System.out.println('Largest sum = ' + largestSum(arr n)); } }
# Python3 implementation of largest # sum contiguous increasing subarray # Returns sum of longest # increasing subarray. def largestSum(arr n): # Initialize result result = -2147483648 # Note that i is incremented # by inner loop also so overall # time complexity is O(n) for i in range(n): # Find sum of longest increasing # subarray starting from arr[i] curr_sum = arr[i] while (i + 1 < n and arr[i + 1] > arr[i]): curr_sum += arr[i + 1] i += 1 # Update result if required if (curr_sum > result): result = curr_sum # required largest sum return result # Driver Code arr = [1 1 4 7 3 6] n = len(arr) print('Largest sum = ' largestSum(arr n)) # This code is contributed by Anant Agarwal.
// C# implementation of largest sum // contiguous increasing subarray using System; class GFG { // Returns sum of longest // increasing subarray. static int largestSum(int[] arr int n) { // Initialize result int result = -9999999; // Note that i is incremented by // inner loop also so overall // time complexity is O(n) for (int i = 0; i < n; i++) { // Find sum of longest increasing // subarray starting from arr[i] int curr_sum = arr[i]; while (i + 1 < n && arr[i + 1] > arr[i]) { curr_sum += arr[i + 1]; i++; } // Update result if required if (curr_sum > result) result = curr_sum; } // required largest sum return result; } // Driver code public static void Main() { int[] arr = { 1 1 4 7 3 6 }; int n = arr.Length; Console.Write('Largest sum = ' + largestSum(arr n)); } } // This code is contributed // by Nitin Mittal.
<script> // Javascript implementation of largest sum // contiguous increasing subarray // Returns sum of longest // increasing subarray. function largestSum(arr n) { // Initialize result var result = -1000000000; // Note that i is incremented // by inner loop also so overall // time complexity is O(n) for (var i = 0; i < n; i++) { // Find sum of longest // increasing subarray // starting from arr[i] var curr_sum = arr[i]; while (i + 1 < n && arr[i + 1] > arr[i]) { curr_sum += arr[i + 1]; i++; } // Update result if required if (curr_sum > result) result = curr_sum; } // required largest sum return result; } // Driver Code var arr = [1 1 4 7 3 6]; var n = arr.length; document.write( 'Largest sum = ' + largestSum(arr n)); // This code is contributed by itsok. </script>
// PHP implementation of largest sum // contiguous increasing subarray // Returns sum of longest // increasing subarray. function largestSum($arr $n) { $INT_MIN = 0; // Initialize result $result = $INT_MIN; // Note that i is incremented // by inner loop also so overall // time complexity is O(n) for ($i = 0; $i < $n; $i++) { // Find sum of longest // increasing subarray // starting from arr[i] $curr_sum = $arr[$i]; while ($i + 1 < $n && $arr[$i + 1] > $arr[$i]) { $curr_sum += $arr[$i + 1]; $i++; } // Update result if required if ($curr_sum > $result) $result = $curr_sum; } // required largest sum return $result; } // Driver Code { $arr = array(1 1 4 7 3 6); $n = sizeof($arr) / sizeof($arr[0]); echo 'Largest sum = ' largestSum($arr $n); return 0; } // This code is contributed by nitin mittal. ?> Išvestis
Largest sum = 12
Laiko sudėtingumas: O(n)
Didžiausia suma gretima didėjanti pogrupis Naudojant Rekursija :
Rekursyvus algoritmas šiai problemai išspręsti:
Štai žingsnis po žingsnio problemos algoritmas:
- Funkcija „didžiausia suma“ užima masyvą "arr" ir jo dydis yra 'n'.
- Jeigu 'n==1' tada grįžti arr[0]th elementas.
- Jeigu 'n != 1' tada rekursyvus funkcijos iškvietimas „didžiausia suma“ rasti didžiausią pogrupio sumą „arr[0...n-1]“ neįskaitant paskutinio elemento „arr[n-1]“ .
- Kartodami per masyvą atvirkštine tvarka, pradedant antruoju paskutiniu elementu, apskaičiuokite didėjančios posistemės sumą, kuri baigiasi „arr[n-1]“ . Jei vienas elementas yra mažesnis už kitą, jis turi būti pridėtas prie esamos sumos. Priešingu atveju kilpa turėtų būti nutraukta.
- Tada grąžinkite didžiausios sumos maksimumą t.y. ' return max(max_sum curr_sum);' .
Čia yra aukščiau pateikto algoritmo įgyvendinimas:
C++#include
#include
/*package whatever //do not write package name here */ import java.util.*; public class Main { // Recursive function to find the largest sum // of contiguous increasing subarray public static int largestSum(int arr[] int n) { // Base case if (n == 1) return arr[0]; // Recursive call to find the largest sum int max_sum = Math.max(largestSum(arr n - 1) arr[n - 1]); // Compute the sum of the increasing subarray int curr_sum = arr[n - 1]; for (int i = n - 2; i >= 0; i--) { if (arr[i] < arr[i + 1]) curr_sum += arr[i]; else break; } // Return the maximum of the largest sum so far // and the sum of the current increasing subarray return Math.max(max_sum curr_sum); } // Driver code public static void main(String[] args) { int arr[] = { 1 1 4 7 3 6 }; int n = arr.length; System.out.println('Largest sum = ' + largestSum(arr n)); } } // This code is contributed by Vaibhav Saroj.
def largestSum(arr n): # Base case if n == 1: return arr[0] # Recursive call to find the largest sum max_sum = max(largestSum(arr n-1) arr[n-1]) # Compute the sum of the increasing subarray curr_sum = arr[n-1] for i in range(n-2 -1 -1): if arr[i] < arr[i+1]: curr_sum += arr[i] else: break # Return the maximum of the largest sum so far # and the sum of the current increasing subarray return max(max_sum curr_sum) # Driver code arr = [1 1 4 7 3 6] n = len(arr) print('Largest sum =' largestSum(arr n)) # This code is contributed by Vaibhav Saroj.
// C# program for above approach using System; public static class GFG { // Recursive function to find the largest sum // of contiguous increasing subarray public static int largestSum(int[] arr int n) { // Base case if (n == 1) return arr[0]; // Recursive call to find the largest sum int max_sum = Math.Max(largestSum(arr n - 1) arr[n - 1]); // Compute the sum of the increasing subarray int curr_sum = arr[n - 1]; for (int i = n - 2; i >= 0; i--) { if (arr[i] < arr[i + 1]) curr_sum += arr[i]; else break; } // Return the maximum of the largest sum so far // and the sum of the current increasing subarray return Math.Max(max_sum curr_sum); } // Driver code public static void Main() { int[] arr = { 1 1 4 7 3 6 }; int n = arr.Length; Console.WriteLine('Largest sum = ' + largestSum(arr n)); } } // This code is contributed by Utkarsh Kumar
function largestSum(arr n) { // Base case if (n == 1) return arr[0]; // Recursive call to find the largest sum let max_sum = Math.max(largestSum(arr n-1) arr[n-1]); // Compute the sum of the increasing subarray let curr_sum = arr[n-1]; for (let i = n-2; i >= 0; i--) { if (arr[i] < arr[i+1]) curr_sum += arr[i]; else break; } // Return the maximum of the largest sum so far // and the sum of the current increasing subarray return Math.max(max_sum curr_sum); } // Driver Code let arr = [1 1 4 7 3 6]; let n = arr.length; console.log('Largest sum = ' + largestSum(arr n));
// Recursive function to find the largest sum // of contiguous increasing subarray function largestSum($arr $n) { // Base case if ($n == 1) return $arr[0]; // Recursive call to find the largest sum $max_sum = max(largestSum($arr $n-1) $arr[$n-1]); // Compute the sum of the increasing subarray $curr_sum = $arr[$n-1]; for ($i = $n-2; $i >= 0; $i--) { if ($arr[$i] < $arr[$i+1]) $curr_sum += $arr[$i]; else break; } // Return the maximum of the largest sum so far // and the sum of the current increasing subarray return max($max_sum $curr_sum); } // Driver Code $arr = array(1 1 4 7 3 6); $n = count($arr); echo 'Largest sum = ' . largestSum($arr $n); ?> Išvestis
Largest sum = 12
Laiko sudėtingumas: O(n^2).
Erdvės sudėtingumas: O(n).
Didžiausia gretimos sumos didėjanti pogrupė Naudojant Kadane algoritmą:
Norint gauti didžiausią sumos pogrupį, naudojamas Kadane metodas, tačiau jis daro prielaidą, kad masyve yra ir teigiamų, ir neigiamų reikšmių. Šiuo atveju turime pakeisti algoritmą, kad jis veiktų tik gretimose kylančiose posistemėse.
Štai kaip galime modifikuoti Kadane'o algoritmą, kad surastume didžiausią gretimos didėjančios pogrupio sumą:
- Pirmajam masyvo elementui inicijuokite du kintamuosius: max_sum ir curr_sum.
- Pereikite per masyvą, pradedant nuo antrojo elemento.
- jei dabartinis elementas yra didesnis nei ankstesnis elementas, pridėkite jį prie curr_sum. Kitu atveju iš naujo nustatykite curr_sum į dabartinį elementą.
- Jei curr_sum yra didesnis nei max_sum, atnaujinkite max_sum.
- Po ciklo max_sum bus didžiausia gretimos didėjančios pogrupio suma.
#include
public class Main { public static int largestSumContiguousIncreasingSubarray(int[] arr int n) { int maxSum = arr[0]; int currSum = arr[0]; for (int i = 1; i < n; i++) { if (arr[i] > arr[i-1]) { currSum += arr[i]; } else { currSum = arr[i]; } if (currSum > maxSum) { maxSum = currSum; } } return maxSum; } public static void main(String[] args) { int[] arr = { 1 1 4 7 3 6 }; int n = arr.length; System.out.println(largestSumContiguousIncreasingSubarray(arr n)); // Output: 44 (1+2+3+5+7+8+9+10) } }
def largest_sum_contiguous_increasing_subarray(arr n): max_sum = arr[0] curr_sum = arr[0] for i in range(1 n): if arr[i] > arr[i-1]: curr_sum += arr[i] else: curr_sum = arr[i] if curr_sum > max_sum: max_sum = curr_sum return max_sum arr = [1 1 4 7 3 6] n = len(arr) print(largest_sum_contiguous_increasing_subarray(arr n)) #output 12 (1+4+7)
using System; class GFG { // Function to find the largest sum of a contiguous // increasing subarray static int LargestSumContiguousIncreasingSubarray(int[] arr int n) { int maxSum = arr[0]; // Initialize the maximum sum // and current sum int currSum = arr[0]; for (int i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) // Check if the current // element is greater than the // previous element { currSum += arr[i]; // If increasing add the // element to the current sum } else { currSum = arr[i]; // If not increasing start a // new increasing subarray // from the current element } if (currSum > maxSum) // Update the maximum sum if the // current sum is greater { maxSum = currSum; } } return maxSum; } static void Main() { int[] arr = { 1 1 4 7 3 6 }; int n = arr.Length; Console.WriteLine( LargestSumContiguousIncreasingSubarray(arr n)); } } // This code is contributed by akshitaguprzj3
// Javascript code for above approach // Function to find the largest sum of a contiguous // increasing subarray function LargestSumContiguousIncreasingSubarray(arr n) { let maxSum = arr[0]; // Initialize the maximum sum // and current sum let currSum = arr[0]; for (let i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) // Check if the current // element is greater than the // previous element { currSum += arr[i]; // If increasing add the // element to the current sum } else { currSum = arr[i]; // If not increasing start a // new increasing subarray // from the current element } if (currSum > maxSum) // Update the maximum sum if the // current sum is greater { maxSum = currSum; } } return maxSum; } let arr = [ 1 1 4 7 3 6 ]; let n = arr.length; console.log(LargestSumContiguousIncreasingSubarray(arr n)); // This code is contributed by Pushpesh Raj
Išvestis
12
Laiko sudėtingumas: O(n).
Erdvės sudėtingumas: O(1).
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