Vi har en matrix af heltal, og vi skal finde to sådanne elementer i matrixen, således at summen af disse to elementer er lig med summen af resten af elementer i matrixen.
Eksempler:
Input : arr[] = {2 11 5 1 4 7} Output : Elements are 4 and 11 Note that 4 + 11 = 2 + 5 + 1 + 7 Input : arr[] = {2 4 2 1 11 15} Output : Elements do not exist EN enkel løsning er at overveje hvert par en efter en at finde sin sum og sammenligne summen med summen af resten af elementerne. Hvis vi finder et par, hvis sum er lig med resten af elementer, udskriver vi parret og returnerer sandt. Tidskompleksiteten af denne løsning er O(n3)
An effektiv løsning er at finde summen af alle array-elementer. Lad denne sum være 'sum'. Nu reduceres opgaven til at finde et par med sum lig med sum/2.
En anden optimering er, at et par kun kan eksistere, hvis summen af hele array er lige, fordi vi grundlæggende deler det i to dele med lige stor sum.
- Find summen af hele array. Lad denne sum være 'sum'
- Hvis summen er ulige returneres falsk.
- Find et par med sum lig med 'sum/2' ved at bruge den diskuterede hashing-baserede metode her som metode 2. Hvis et par findes, udskriv det og returner sandt.
- Hvis der ikke findes et par, returneres falsk.
Nedenfor er implementeringen af ovenstående trin.
C++// C++ program to find whether two elements exist // whose sum is equal to sum of rest of the elements. #include using namespace std; // Function to check whether two elements exist // whose sum is equal to sum of rest of the elements. bool checkPair(int arr[] int n) { // Find sum of whole array int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; // If sum of array is not even then we can not // divide it into two part if (sum % 2 != 0) return false; sum = sum / 2; // For each element arr[i] see if there is // another element with value sum - arr[i] unordered_set<int> s; for (int i = 0; i < n; i++) { int val = sum - arr[i]; // If element exist than return the pair if (s.find(val) != s.end()) { printf('Pair elements are %d and %dn' arr[i] val); return true; } s.insert(arr[i]); } return false; } // Driver program. int main() { int arr[] = { 2 11 5 1 4 7 }; int n = sizeof(arr) / sizeof(arr[0]); if (checkPair(arr n) == false) printf('No pair found'); return 0; }
Java // Java program to find whether two elements exist // whose sum is equal to sum of rest of the elements. import java.util.*; class GFG { // Function to check whether two elements exist // whose sum is equal to sum of rest of the elements. static boolean checkPair(int arr[] int n) { // Find sum of whole array int sum = 0; for (int i = 0; i < n; i++) { sum += arr[i]; } // If sum of array is not even then we can not // divide it into two part if (sum % 2 != 0) { return false; } sum = sum / 2; // For each element arr[i] see if there is // another element with value sum - arr[i] HashSet<Integer> s = new HashSet<Integer>(); for (int i = 0; i < n; i++) { int val = sum - arr[i]; // If element exist than return the pair if (s.contains(val) && val == (int)s.toArray()[s.size() - 1]) { System.out.printf( 'Pair elements are %d and %dn' arr[i] val); return true; } s.add(arr[i]); } return false; } // Driver program. public static void main(String[] args) { int arr[] = { 2 11 5 1 4 7 }; int n = arr.length; if (checkPair(arr n) == false) { System.out.printf('No pair found'); } } } /* This code contributed by PrinciRaj1992 */
Python3 # Python3 program to find whether # two elements exist whose sum is # equal to sum of rest of the elements. # Function to check whether two # elements exist whose sum is equal # to sum of rest of the elements. def checkPair(arr n): s = set() sum = 0 # Find sum of whole array for i in range(n): sum += arr[i] # / If sum of array is not # even then we can not # divide it into two part if sum % 2 != 0: return False sum = sum / 2 # For each element arr[i] see if # there is another element with # value sum - arr[i] for i in range(n): val = sum - arr[i] if arr[i] not in s: s.add(arr[i]) # If element exist than # return the pair if val in s: print('Pair elements are' arr[i] 'and' int(val)) # Driver Code arr = [2 11 5 1 4 7] n = len(arr) if checkPair(arr n) == False: print('No pair found') # This code is contributed # by Shrikant13
C# // C# program to find whether two elements exist // whose sum is equal to sum of rest of the elements. using System; using System.Collections.Generic; class GFG { // Function to check whether two elements exist // whose sum is equal to sum of rest of the elements. static bool checkPair(int []arr int n) { // Find sum of whole array int sum = 0; for (int i = 0; i < n; i++) { sum += arr[i]; } // If sum of array is not even then we can not // divide it into two part if (sum % 2 != 0) { return false; } sum = sum / 2; // For each element arr[i] see if there is // another element with value sum - arr[i] HashSet<int> s = new HashSet<int>(); for (int i = 0; i < n; i++) { int val = sum - arr[i]; // If element exist than return the pair if (s.Contains(val)) { Console.Write('Pair elements are {0} and {1}n' arr[i] val); return true; } s.Add(arr[i]); } return false; } // Driver code public static void Main(String[] args) { int []arr = {2 11 5 1 4 7}; int n = arr.Length; if (checkPair(arr n) == false) { Console.Write('No pair found'); } } } // This code contributed by Rajput-Ji
PHP // PHP program to find whether two elements exist // whose sum is equal to sum of rest of the elements. // Function to check whether two elements exist // whose sum is equal to sum of rest of the elements. function checkPair(&$arr $n) { // Find sum of whole array $sum = 0; for ($i = 0; $i < $n; $i++) $sum += $arr[$i]; // If sum of array is not even then we // can not divide it into two part if ($sum % 2 != 0) return false; $sum = $sum / 2; // For each element arr[i] see if there is // another element with value sum - arr[i] $s = array(); for ($i = 0; $i < $n; $i++) { $val = $sum - $arr[$i]; // If element exist than return the pair if (array_search($val $s)) { echo 'Pair elements are ' . $arr[$i] . ' and ' . $val . 'n'; return true; } array_push($s $arr[$i]); } return false; } // Driver Code $arr = array(2 11 5 1 4 7); $n = sizeof($arr); if (checkPair($arr $n) == false) echo 'No pair found'; // This code is contributed by ita_c ?> JavaScript <script> // Javascript program to find // whether two elements exist // whose sum is equal to sum of rest // of the elements. // Function to check whether // two elements exist // whose sum is equal to sum of // rest of the elements. function checkPair(arrn) { // Find sum of whole array let sum = 0; for (let i = 0; i < n; i++) { sum += arr[i]; } // If sum of array is not even then we can not // divide it into two part if (sum % 2 != 0) { return false; } sum = Math.floor(sum / 2); // For each element arr[i] see if there is // another element with value sum - arr[i] let s = new Set(); for (let i = 0; i < n; i++) { let val = sum - arr[i]; // If element exist than return the pair if(!s.has(arr[i])) { s.add(arr[i]) } if (s.has(val) ) { document.write('Pair elements are '+ arr[i]+' and '+ val+'
'); return true; } s.add(arr[i]); } return false; } // Driver program. let arr=[2 11 5 1 4 7]; let n = arr.length; if (checkPair(arr n) == false) { document.write('No pair found'); } // This code is contributed by rag2127 </script>
Produktion
Pair elements are 4 and 11
Tidskompleksitet: På) . unordered_set implementeres ved hjælp af hashing. Tidskompleksitet hashsøgning og indsættelse antages som O(1) her.
Hjælpeplads: På)
En anden effektiv tilgang (pladsoptimering): Først vil vi sortere arrayet efter Binær søgning . Så vil vi iterere hele arrayet og kontrollere, om der findes et indeks i arrayet, der parrer med i, således at arr[indeks] + a[i] == Restsummen af arrayet. Vi kan bruge binær søgning til at finde et indeks i arrayet ved at ændre det binære søgeprogram. Hvis der findes et par, så udskriv det par. ellers print findes der intet par.
Nedenfor er implementeringen af ovenstående tilgang:
C++// C++ program for the above approach #include using namespace std; // Function to Find if a index exist in array such that // arr[index] + a[i] == Rest sum of the array int binarysearch(int arr[] int n int i int Totalsum) { int l = 0 r = n - 1 index = -1;//initialize as -1 while (l <= r) { int mid = (l + r) / 2; int Pairsum = arr[mid] + arr[i];//pair sum int Restsum = Totalsum - Pairsum;//Rest sum if ( Pairsum == Restsum ) { if( index != i )// checking a pair has same position or not { index = mid; }//Then update index -1 to mid // Checking for adjacent element else if(index == i && mid>0 && arr[mid-1]==arr[i]) { index = mid-1; }//Then update index -1 to mid-1 else if(index == i && mid<n-1 && arr[mid+1]==arr[i]) { index = mid+1; } //Then update index-1 to mid+1 break; } else if (Pairsum > Restsum) { // If pair sum is greater than rest sum our index will // be in the Range [mid+1R] l = mid + 1; } else { // If pair sum is smaller than rest sum our index will // be in the Range [Lmid-1] r = mid - 1; } } // return index=-1 if a pair not exist with arr[i] // else return index such that arr[i]+arr[index] == sum of rest of arr return index; } // Function to check if a pair exist such their sum // equal to rest of the array or not bool checkPair(int arr[]int n) { int Totalsum=0; sort(arr arr + n);//sort arr for Binary search for(int i=0;i<n;i++) { Totalsum+=arr[i]; } //Finding total sum of the arr for(int i=0;i<n;i++) { // If index is -1 Means arr[i] can't pair with any element // else arr[i]+a[index] == sum of rest of the arr int index = binarysearch(arr n iTotalsum) ; if(index != -1) { cout<<'Pair elements are '<< arr[i]<<' and '<< arr[index]; return true; } } return false;//Return false if a pair not exist } // Driver Code int main() { int arr[] = {2 11 5 1 4 7}; int n = sizeof(arr)/sizeof(arr[0]); //Function call if (checkPair(arr n) == false) { cout<<'No pair found'; } return 0; } // This Approach is contributed by nikhilsainiofficial546
Java // Java program for the above approach import java.util.*; class GFG { // Function to Find if a index exist in array such that // arr[index] + a[i] == Rest sum of the array static int binarysearch(int arr[] int n int i int Totalsum) { int l = 0 r = n - 1 index = -1; // initialize as -1 while (l <= r) { int mid = (l + r) / 2; int Pairsum = arr[mid] + arr[i]; // pair sum int Restsum = Totalsum - Pairsum; // Rest sum if (Pairsum == Restsum) { if (index != i) // checking a pair has same // position or not { index = mid; } // Then update index -1 to mid // Checking for adjacent element else if (index == i && mid > 0 && arr[mid - 1] == arr[i]) { index = mid - 1; } // Then update index -1 to mid-1 else if (index == i && mid < n - 1 && arr[mid + 1] == arr[i]) { index = mid + 1; } // Then update index-1 to mid+1 break; } else if (Pairsum > Restsum) { // If pair sum is greater than rest sum // our index will be in the Range [mid+1R] l = mid + 1; } else { // If pair sum is smaller than rest sum // our index will be in the Range [Lmid-1] r = mid - 1; } } // return index=-1 if a pair not exist with arr[i] // else return index such that arr[i]+arr[index] == // sum of rest of arr return index; } // Function to check if a pair exist such their sum // equal to rest of the array or not static boolean checkPair(int arr[] int n) { int Totalsum = 0; Arrays.sort(arr); // sort arr for Binary search for (int i = 0; i < n; i++) { Totalsum += arr[i]; } // Finding total sum of the arr for (int i = 0; i < n; i++) { // If index is -1 Means arr[i] can't pair with // any element else arr[i]+a[index] == sum of // rest of the arr int index = binarysearch(arr n i Totalsum); if (index != -1) { System.out.println('Pair elements are ' + arr[i] + ' and ' + arr[index]); return true; } } return false; // Return false if a pair not exist } // Driver Code public static void main(String[] args) { int arr[] = { 2 11 5 1 4 7 }; int n = arr.length; // Function call if (checkPair(arr n) == false) { System.out.println('No pair found'); } } }
Python3 # Python program for the above approach # Function to find if a index exist in array such that # arr[index] + a[i] == Rest sum of the array def binarysearch(arr n i Totalsum): l = 0 r = n - 1 index = -1 # Initialize as -1 while l <= r: mid = (l + r) // 2 Pairsum = arr[mid] + arr[i] # Pair sum Restsum = Totalsum - Pairsum # Rest sum if Pairsum == Restsum: if index != i: # Checking if a pair has the same position or not index = mid # Then update index -1 to mid # Checking for adjacent element elif index == i and mid > 0 and arr[mid - 1] == arr[i]: index = mid - 1 # Then update index -1 to mid-1 elif index == i and mid < n - 1 and arr[mid + 1] == arr[i]: index = mid + 1 # Then update index-1 to mid+1 break elif Pairsum > Restsum: # If pair sum is greater than rest sum our index will # be in the Range [mid+1R] l = mid + 1 else: # If pair sum is smaller than rest sum our index will # be in the Range [Lmid-1] r = mid - 1 # Return index=-1 if a pair not exist with arr[i] # else return index such that arr[i]+arr[index] == sum of rest of arr return index # Function to check if a pair exists such that their sum # equals to rest of the array or not def checkPair(arr n): Totalsum = 0 arr = sorted(arr) # Sort arr for Binary search for i in range(n): Totalsum += arr[i] # Finding total sum of the arr for i in range(n): # If index is -1 means arr[i] can't pair with any element # else arr[i]+a[index] == sum of rest of the arr index = binarysearch(arr n i Totalsum) if index != -1: print('Pair elements are' arr[i] 'and' arr[index]) return True return False # Return false if a pair not exist # Driver Code arr = [2 11 5 1 4 7] n = len(arr) # Function call if checkPair(arr n) == False: print('No pair found')
C# using System; class GFG { // Function to Find if a index exist in array such that // arr[index] + a[i] == Rest sum of the array static int BinarySearch(int[] arr int n int i int totalSum) { int l = 0 r = n - 1 index = -1; // initialize as -1 while (l <= r) { int mid = (l + r) / 2; int pairSum = arr[mid] + arr[i]; // pair sum int restSum = totalSum - pairSum; // rest sum if (pairSum == restSum) { if (index != i) // checking a pair has same // position or not { index = mid; } // Then update index -1 to mid // Checking for adjacent element else if (index == i && mid > 0 && arr[mid - 1] == arr[i]) { index = mid - 1; } // Then update index -1 to mid-1 else if (index == i && mid < n - 1 && arr[mid + 1] == arr[i]) { index = mid + 1; } // Then update index-1 to mid+1 break; } else if (pairSum > restSum) { // If pair sum is greater than rest sum // our index will be in the Range [mid+1R] l = mid + 1; } else { // If pair sum is smaller than rest sum // our index will be in the Range [Lmid-1] r = mid - 1; } } // return index=-1 if a pair not exist with arr[i] // else return index such that arr[i]+arr[index] == // sum of rest of arr return index; } // Function to check if a pair exist such their sum // equal to rest of the array or not static bool CheckPair(int[] arr int n) { int totalSum = 0; Array.Sort(arr); // sort arr for Binary search for (int i = 0; i < n; i++) { totalSum += arr[i]; } // Finding total sum of the arr for (int i = 0; i < n; i++) { // If index is -1 Means arr[i] can't pair with // any element else arr[i]+a[index] == sum of // rest of the arr int index = BinarySearch(arr n i totalSum); if (index != -1) { Console.WriteLine('Pair elements are ' + arr[i] + ' and ' + arr[index]); return true; } } return false; // Return false if a pair not exist } // Driver Code static void Main(string[] args) { int[] arr = { 2 11 5 1 4 7 }; int n = arr.Length; // Function call if (!CheckPair(arr n)) { Console.WriteLine('No pair found'); } } }
JavaScript // JavaScript program for the above approach // function to find if a index exist in array such that // arr[index] + a[i] == Rest sum of the array function binarysearch(arr n i TotalSum){ let l = 0; let r = n-1; let index = -1; while(l <= r){ let mid = parseInt((l+r)/2); let Pairsum = arr[mid] + arr[i]; let Restsum = TotalSum - Pairsum; if ( Pairsum == Restsum ) { if( index != i )// checking a pair has same position or not { index = mid; }//Then update index -1 to mid // Checking for adjacent element else if(index == i && mid>0 && arr[mid-1]==arr[i]) { index = mid-1; }//Then update index -1 to mid-1 else if(index == i && mid<n-1 && arr[mid+1]==arr[i]) { index = mid+1; } //Then update index-1 to mid+1 break; } else if (Pairsum > Restsum) { // If pair sum is greater than rest sum our index will // be in the Range [mid+1R] l = mid + 1; } else { // If pair sum is smaller than rest sum our index will // be in the Range [Lmid-1] r = mid - 1; } } // return index=-1 if a pair not exist with arr[i] // else return index such that arr[i]+arr[index] == sum of rest of arr return index; } // Function to check if a pair exist such their sum // equal to rest of the array or not function checkPair(arr n){ let Totalsum = 0; arr.sort(function(a b){return a - b}); for(let i=0;i<n;i++) { Totalsum+=arr[i]; } //Finding total sum of the arr for(let i=0;i<n;i++) { // If index is -1 Means arr[i] can't pair with any element // else arr[i]+a[index] == sum of rest of the arr let index = binarysearch(arr n iTotalsum) ; if(index != -1) { console.log('Pair elements are ' + arr[i] + ' and ' + arr[index]); return true; } } return false;//Return false if a pair not exist } // driver code to test above function let arr = [2 11 5 1 4 7]; let n = arr.length; // function call if(checkPair(arr n) == false) console.log('No Pair Found') // THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)
Produktion
Pair elements are 11 and 4
Tidskompleksitet: O(n * logn)
Hjælpeplads: O(1)
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