#працтицеЛинкДив { дисплаи: ноне !импортант; }Дат низ целих бројева и број к. Можемо упарити два броја низа ако је разлика између њих стриктно мања од к. Задатак је пронаћи највећи могући збир дисјунктних парова. Збир П парова је збир свих 2П бројева парова.
Примери:
Препоручена пракса Парови са специфичном разликом Покушајте!Унос : арр[] = {3 5 10 15 17 12 9} К = 4
Излаз : 62
Објашњење:
Тада су дисјунктни парови са разликом мањом од К (3 5) (10 12) (15 17)
Дакле, максимални збир који можемо добити је 3 + 5 + 12 + 10 + 15 + 17 = 62
Имајте на уму да је алтернативни начин формирања дисјунктних парова (3 5) (9 12) (15 17), али ово упаривање производи мањи збир.
Унос : арр[] = {5 15 10 300} к = 12
Излаз : 25
приступ: Прво сортирамо дати низ по растућем редоследу. Када је низ сортиран, прелазимо низ низ. За сваки елемент покушавамо да га прво упаримо са претходним елементом. Зашто преферирамо претходни елемент? Нека арр[и] може бити упарен са арр[и-1] и арр[и-2] (тј. арр[и] – арр[и-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.
Сада посматрајући горе наведене чињенице можемо формулисати наше решење за динамичко програмирање као у наставку
Нека дп[и] означава максималну суму дисјунктних парова коју можемо постићи коришћењем првих и елемената низа. Претпоставимо да смо тренутно на и-ој позицији, онда постоје две могућности за нас.
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]
Горенаведена итерација траје О(Н) времена, а сортирање низа ће трајати О(Н лог Н) времена тако да ће укупна временска сложеност решења бити О(Н лог Н)
Имплементација:
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>
Излаз
62
Временска сложеност: О(Н Лог Н)
Помоћни простор: О(Н)
Оптимизовано решење које је допринео Амит Сане је дато у наставку
Имплементација:
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>
Излаз
62
Временска сложеност: О(Н Лог Н)
Помоћни простор: О(1)