Maximal summa av par med specifik skillnad
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Produktion
Produktion
#practiceLinkDiv { display: ingen !viktigt; } Givet en matris med heltal och ett tal k. Vi kan para ihop två nummer i matrisen om skillnaden mellan dem är strikt mindre än k. Uppgiften är att hitta den maximala möjliga summan av disjunkta par. Summan av P-par är summan av alla 2P-antal av par.
Exempel:
Rekommenderad praxis Par med specifik skillnad Prova det!Inmatning : arr[] = {3 5 10 15 17 12 9} K = 4
Utgång: 62
Förklaring:
Då är osammanhängande par med skillnad mindre än K (3 5) (10 12) (15 17)
Så den maximala summan vi kan få är 3 + 5 + 12 + 10 + 15 + 17 = 62
Observera att ett alternativt sätt att bilda disjunkta par är (3 5) (9 12) (15 17) men denna parning ger mindre summa.
Inmatning : arr[] = {5 15 10 300} k = 12
Utgång: 25
Närma sig: Först sorterar vi den givna matrisen i ökande ordning. När arrayen är sorterad passerar vi arrayen. För varje element försöker vi först para ihop det med dess tidigare element. Varför föredrar vi tidigare element? Låt arr[i] kan paras ihop med arr[i-1] och arr[i-2] (dvs. 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.
När vi nu observerar ovanstående fakta kan vi formulera vår dynamiska programmeringslösning enligt nedan
Låt dp[i] betecknar den maximala disjunkta parsumman som vi kan uppnå genom att använda första i-elementen i arrayen. Anta att vi för närvarande är på i'th position så finns det två möjligheter för oss.
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]
Ovan iteration tar O(N) tid och sortering av array kommer att ta O(N log N) tid så den totala tidskomplexiteten för lösningen blir O(N log N)
Genomförande:
C++ // C++ program to find maximum pair sum whose // difference is less than K #include using namespace std ; // method to return maximum sum we can get by // finding less than K difference pair int maxSumPairWithDifferenceLessThanK ( int arr [] int N int K ) { // Sort input array in ascending order. sort ( arr arr + N ); // dp[i] denotes the maximum disjoint pair sum // we can achieve using first i elements int dp [ 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 ] = 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 int main () { int arr [] = { 3 5 10 15 17 12 9 }; int N = sizeof ( arr ) / sizeof ( int ); int K = 4 ; cout < < maxSumPairWithDifferenceLessThanK ( arr N K ); return 0 ; }
Java // 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 # 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# // 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 // 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 ?>
JavaScript < 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>
Produktion
62
Tidskomplexitet: O(N Log N)
Hjälputrymme: O(N)
En optimerad lösning med bidrag från Amit Sane ges nedan
Genomförande:
C++ // C++ program to find maximum pair sum whose // difference is less than K #include using namespace std ; // Method to return maximum sum we can get by // finding less than K difference pairs int maxSumPair ( int arr [] int N int k ) { int maxSum = 0 ; // Sort elements to ensure every i and i-1 is closest // possible pair sort ( arr arr + N ); // 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 Kadd 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] < // rr[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 int main () { int arr [] = { 3 5 10 15 17 12 9 }; int N = sizeof ( arr ) / sizeof ( int ); int K = 4 ; cout < < maxSumPair ( arr N K ); return 0 ; }
Java // 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 # 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# // 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 // 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 ?>
JavaScript < 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 (( a b )=> 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>
Produktion
62
Tidskomplexitet: O(N Log N)
Hjälputrymme: O(1)
