#practiceLinkDiv { display: none !important; }Podana je matrika celih števil in število k. Dve števili niza lahko seznanimo, če je razlika med njima strogo manjša od k. Naloga je najti največjo možno vsoto disjunktnih parov. Vsota P parov je vsota vseh 2P števil parov.
Primeri:
Priporočena praksa Pari s posebno razliko Poskusite!Vnos : arr[] = {3 5 10 15 17 12 9} K = 4
Izhod: 62
Pojasnilo:
Potem so disjunktni pari z razliko manjšo od K (3 5) (10 12) (15 17)
Torej je največja vsota, ki jo lahko dobimo, 3 + 5 + 12 + 10 + 15 + 17 = 62
Upoštevajte, da je alternativni način za oblikovanje disjunktnih parov (3 5) (9 12) (15 17), vendar to združevanje povzroči manjšo vsoto.
Vnos : arr[] = {5 15 10 300} k = 12
Izhod: 25
Pristop: Najprej razvrstimo dano matriko v naraščajočem vrstnem redu. Ko je matrika razvrščena, jo prečkamo. Za vsak element ga poskušamo najprej seznaniti s prejšnjim elementom. Zakaj imamo raje prejšnji element? Naj se arr[i] seznani z arr[i-1] in arr[i-2] (tj. arr[i] – arr[i-1]< K and arr[i]-arr[i-2] < K). Since the array is sorted value of arr[i-1] would be more than arr[i-2]. Also we need to pair with difference less than k it means if arr[i-2] can be paired then arr[i-1] can also be paired in a sorted array.
Ob upoštevanju zgornjih dejstev lahko oblikujemo našo rešitev za dinamično programiranje, kot je prikazano spodaj
Naj dp[i] označuje največjo disjonktno vsoto parov, ki jo lahko dosežemo z uporabo prvih i elementov matrike. Predpostavimo, da smo trenutno na i-ti poziciji, potem imamo dve možnosti.
Pair up i with (i-1)th element i.e. dp[i] = dp[i-2] + arr[i] + arr[i-1] Don't pair up i.e. dp[i] = dp[i-1]
Zgornja iteracija traja O(N) časa in razvrščanje matrike bo trajalo O(N log N) časa, tako da bo skupna časovna kompleksnost rešitve O(N log N)
Izvedba:
C++// C++ program to find maximum pair sum whose // difference is less than K #include
// Java program to find maximum pair sum whose // difference is less than K import java.io.*; import java.util.*; class GFG { // method to return maximum sum we can get by // finding less than K difference pair static int maxSumPairWithDifferenceLessThanK(int arr[] int N int K) { // Sort input array in ascending order. Arrays.sort(arr); // dp[i] denotes the maximum disjoint pair sum // we can achieve using first i elements int dp[] = new int[N]; // if no element then dp value will be 0 dp[0] = 0; for (int i = 1; i < N; i++) { // first give previous value to dp[i] i.e. // no pairing with (i-1)th element dp[i] = dp[i-1]; // if current and previous element can form a pair if (arr[i] - arr[i-1] < K) { // update dp[i] by choosing maximum between // pairing and not pairing if (i >= 2) dp[i] = Math.max(dp[i] dp[i-2] + arr[i] + arr[i-1]); else dp[i] = Math.max(dp[i] arr[i] + arr[i-1]); } } // last index will have the result return dp[N - 1]; } // Driver code to test above methods public static void main (String[] args) { int arr[] = {3 5 10 15 17 12 9}; int N = arr.length; int K = 4; System.out.println ( maxSumPairWithDifferenceLessThanK( arr N K)); } } //This code is contributed by vt_m.
# Python3 program to find maximum pair # sum whose difference is less than K # method to return maximum sum we can # get by get by finding less than K # difference pair def maxSumPairWithDifferenceLessThanK(arr N K): # Sort input array in ascending order. arr.sort() # dp[i] denotes the maximum disjoint # pair sum we can achieve using first # i elements dp = [0] * N # if no element then dp value will be 0 dp[0] = 0 for i in range(1 N): # first give previous value to # dp[i] i.e. no pairing with # (i-1)th element dp[i] = dp[i-1] # if current and previous element # can form a pair if (arr[i] - arr[i-1] < K): # update dp[i] by choosing # maximum between pairing # and not pairing if (i >= 2): dp[i] = max(dp[i] dp[i-2] + arr[i] + arr[i-1]); else: dp[i] = max(dp[i] arr[i] + arr[i-1]); # last index will have the result return dp[N - 1] # Driver code to test above methods arr = [3 5 10 15 17 12 9] N = len(arr) K = 4 print(maxSumPairWithDifferenceLessThanK(arr N K)) # This code is contributed by Smitha Dinesh Semwal
// C# program to find maximum pair sum whose // difference is less than K using System; class GFG { // method to return maximum sum we can get by // finding less than K difference pair static int maxSumPairWithDifferenceLessThanK(int []arr int N int K) { // Sort input array in ascending order. Array.Sort(arr); // dp[i] denotes the maximum disjoint pair sum // we can achieve using first i elements int []dp = new int[N]; // if no element then dp value will be 0 dp[0] = 0; for (int i = 1; i < N; i++) { // first give previous value to dp[i] i.e. // no pairing with (i-1)th element dp[i] = dp[i-1]; // if current and previous element can form // a pair if (arr[i] - arr[i-1] < K) { // update dp[i] by choosing maximum // between pairing and not pairing if (i >= 2) dp[i] = Math.Max(dp[i] dp[i-2] + arr[i] + arr[i-1]); else dp[i] = Math.Max(dp[i] arr[i] + arr[i-1]); } } // last index will have the result return dp[N - 1]; } // Driver code to test above methods public static void Main () { int []arr = {3 5 10 15 17 12 9}; int N = arr.Length; int K = 4; Console.WriteLine( maxSumPairWithDifferenceLessThanK(arr N K)); } } // This code is contributed by anuj_67.
// Php program to find maximum pair sum whose // difference is less than K // method to return maximum sum we can get by // finding less than K difference pair function maxSumPairWithDifferenceLessThanK($arr $N $K) { // Sort input array in ascending order. sort($arr) ; // dp[i] denotes the maximum disjoint pair sum // we can achieve using first i elements $dp = array() ; // if no element then dp value will be 0 $dp[0] = 0; for ($i = 1; $i < $N; $i++) { // first give previous value to dp[i] i.e. // no pairing with (i-1)th element $dp[$i] = $dp[$i-1]; // if current and previous element can form a pair if ($arr[$i] - $arr[$i-1] < $K) { // update dp[i] by choosing maximum between // pairing and not pairing if ($i >= 2) $dp[$i] = max($dp[$i] $dp[$i-2] + $arr[$i] + $arr[$i-1]); else $dp[$i] = max($dp[$i] $arr[$i] + $arr[$i-1]); } } // last index will have the result return $dp[$N - 1]; } // Driver code $arr = array(3 5 10 15 17 12 9); $N = sizeof($arr) ; $K = 4; echo maxSumPairWithDifferenceLessThanK($arr $N $K); // This code is contributed by Ryuga ?> <script> // Javascript program to find maximum pair sum whose // difference is less than K // method to return maximum sum we can get by // finding less than K difference pair function maxSumPairWithDifferenceLessThanK(arr N K) { // Sort input array in ascending order. arr.sort(); // dp[i] denotes the maximum disjoint pair sum // we can achieve using first i elements let dp = []; // if no element then dp value will be 0 dp[0] = 0; for (let i = 1; i < N; i++) { // first give previous value to dp[i] i.e. // no pairing with (i-1)th element dp[i] = dp[i-1]; // if current and previous element can form a pair if (arr[i] - arr[i-1] < K) { // update dp[i] by choosing maximum between // pairing and not pairing if (i >= 2) dp[i] = Math.max(dp[i] dp[i-2] + arr[i] + arr[i-1]); else dp[i] = Math.max(dp[i] arr[i] + arr[i-1]); } } // last index will have the result return dp[N - 1]; } // Driver code to test above methods let arr = [3 5 10 15 17 12 9]; let N = arr.length; let K = 4; document.write( maxSumPairWithDifferenceLessThanK( arr N K)); // This code is contributed by avijitmondal1998. </script>
Izhod
62
Časovna zahtevnost: O(N Log N)
Pomožni prostor: O(N)
Spodaj je podana optimizirana rešitev, ki jo je prispeval Amit Sane
Izvedba:
C++// C++ program to find maximum pair sum whose // difference is less than K #include
// Java program to find maximum pair sum whose // difference is less than K import java.io.*; import java.util.*; class GFG { // Method to return maximum sum we can get by // finding less than K difference pairs static int maxSumPairWithDifferenceLessThanK(int arr[] int N int k) { int maxSum = 0; // Sort elements to ensure every i and i-1 is // closest possible pair Arrays.sort(arr); // To get maximum possible sum // iterate from largest // to smallest giving larger // numbers priority over // smaller numbers. for (int i = N - 1; i > 0; --i) { // Case I: Diff of arr[i] and arr[i-1] is less // than K add to maxSum // Case II: Diff between arr[i] and arr[i-1] is // not less than K move to next i // since with sorting we know arr[i]-arr[i-1] < // arr[i]-arr[i-2] and so on. if (arr[i] - arr[i - 1] < k) { // Assuming only positive numbers. maxSum += arr[i]; maxSum += arr[i - 1]; // When a match is found skip this pair --i; } } return maxSum; } // Driver code public static void main(String[] args) { int arr[] = { 3 5 10 15 17 12 9 }; int N = arr.length; int K = 4; System.out.println( maxSumPairWithDifferenceLessThanK(arr N K)); } } // This code is contributed by vt_m.
# Python3 program to find maximum pair sum # whose difference is less than K # Method to return maximum sum we can # get by finding less than K difference # pairs def maxSumPairWithDifferenceLessThanK(arr N k): maxSum = 0 # Sort elements to ensure every i and # i-1 is closest possible pair arr.sort() # To get maximum possible sum iterate # from largest to smallest giving larger # numbers priority over smaller numbers. i = N - 1 while (i > 0): # Case I: Diff of arr[i] and arr[i-1] # is less than K add to maxSum # Case II: Diff between arr[i] and # arr[i-1] is not less than K # move to next i since with sorting # we know arr[i]-arr[i-1] < arr[i]-arr[i-2] # and so on. if (arr[i] - arr[i - 1] < k): # Assuming only positive numbers. maxSum += arr[i] maxSum += arr[i - 1] # When a match is found skip this pair i -= 1 i -= 1 return maxSum # Driver Code arr = [3 5 10 15 17 12 9] N = len(arr) K = 4 print(maxSumPairWithDifferenceLessThanK(arr N K)) # This code is contributed by mits
// C# program to find maximum pair sum whose // difference is less than K using System; class GFG { // Method to return maximum sum we can get by // finding less than K difference pairs static int maxSumPairWithDifferenceLessThanK(int []arr int N int k) { int maxSum = 0; // Sort elements to ensure // every i and i-1 is closest // possible pair Array.Sort(arr); // To get maximum possible sum // iterate from largest // to smallest giving larger // numbers priority over // smaller numbers. for (int i = N-1; i > 0; --i) { /* Case I: Diff of arr[i] and arr[i-1] is less than K add to maxSum Case II: Diff between arr[i] and arr[i-1] is not less than K move to next i since with sorting we know arr[i]-arr[i-1] < arr[i]-arr[i-2] and so on.*/ if (arr[i] - arr[i-1] < k) { // Assuming only positive numbers. maxSum += arr[i]; maxSum += arr[i - 1]; // When a match is found // skip this pair --i; } } return maxSum; } // Driver Code public static void Main () { int []arr = {3 5 10 15 17 12 9}; int N = arr.Length; int K = 4; Console.Write( maxSumPairWithDifferenceLessThanK(arr N K)); } } // This code is contributed by nitin mittal.
// PHP program to find maximum pair sum // whose difference is less than K // Method to return maximum sum we can // get by finding less than K difference // pairs function maxSumPairWithDifferenceLessThanK($arr $N $k) { $maxSum = 0; // Sort elements to ensure every i and // i-1 is closest possible pair sort($arr); // To get maximum possible sum iterate // from largest to smallest giving larger // numbers priority over smaller numbers. for ($i = $N - 1; $i > 0; --$i) { // Case I: Diff of arr[i] and arr[i-1] // is less than K add to maxSum // Case II: Diff between arr[i] and // arr[i-1] is not less than K // move to next i since with sorting // we know arr[i]-arr[i-1] < arr[i]-arr[i-2] // and so on. if ($arr[$i] - $arr[$i - 1] < $k) { // Assuming only positive numbers. $maxSum += $arr[$i]; $maxSum += $arr[$i - 1]; // When a match is found skip this pair --$i; } } return $maxSum; } // Driver Code $arr = array(3 5 10 15 17 12 9); $N = sizeof($arr); $K = 4; echo maxSumPairWithDifferenceLessThanK($arr $N $K); // This code is contributed // by Sach_Code ?> <script> // Javascript program to find // maximum pair sum whose // difference is less than K // Method to return maximum sum we can get by // finding less than K difference pairs function maxSumPairWithDifferenceLessThanK(arr N k) { var maxSum = 0; // Sort elements to ensure every i and i-1 is // closest possible pair arr.sort((ab)=>a-b); // To get maximum possible sum // iterate from largest // to smallest giving larger // numbers priority over // smaller numbers. for (i = N - 1; i > 0; --i) { // Case I: Diff of arr[i] and arr[i-1] is less // than K add to maxSum // Case II: Diff between arr[i] and arr[i-1] is // not less than K move to next i // since with sorting we know arr[i]-arr[i-1] < // arr[i]-arr[i-2] and so on. if (arr[i] - arr[i - 1] < k) { // Assuming only positive numbers. maxSum += arr[i]; maxSum += arr[i - 1]; // When a match is found skip this pair --i; } } return maxSum; } // Driver code var arr = [ 3 5 10 15 17 12 9 ]; var N = arr.length; var K = 4; document.write(maxSumPairWithDifferenceLessThanK(arr N K)); // This code is contributed by 29AjayKumar </script>
Izhod
62
Časovna zahtevnost: O(N Log N)
Pomožni prostor: O(1)