Имамо низ целих бројева и морамо да пронађемо два таква елемента у низу тако да је збир ова два елемента једнак збиру остатака елемената у низу.
Примери:
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 А једноставно решење је размотрити сваки пар један по један пронаћи његов збир и упоредити збир са збиром осталих елемената. Ако пронађемо пар чији је збир једнак остатку елемената, исписујемо пар и враћамо труе. Временска сложеност овог решења је О(н3)
Ан ефикасно решење је пронаћи збир свих елемената низа. Нека овај збир буде 'збир'. Сада се задатак своди на проналажење пара чији је збир једнак збиру/2.
Друга оптимизација је да пар може постојати само ако је збир целог низа паран јер га у основи делимо на два дела са једнаким збиром.
- Пронађите збир целог низа. Нека овај збир буде 'збир'
- Ако је збир непаран, вратите нетачно.
- Пронађите пар са сумом једнаким 'суму/2' користећи методу засновану на хеширању овде као метод 2. Ако је пар пронађен, одштампајте га и вратите труе.
- Ако ниједан пар не постоји, вратите фалсе.
Испод је имплементација горњих корака.
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>
Излаз
Pair elements are 4 and 11
Временска сложеност: О(н) . унордеред_сет се имплементира помоћу хеширања. Временска сложеност хеш претрага и уметања се овде претпоставља као О(1).
Помоћни простор: О(н)
Још један ефикасан приступ (оптимизација простора): Прво ћемо сортирати низ за Бинарно претраживање . Затим ћемо поновити цео низ и проверити да ли постоји индекс у низу који је упарен са и тако да је арр[индек] + а[и] == Остала сума низа. Можемо да користимо бинарну претрагу да пронађемо индекс у низу тако што ћемо модификовати програм бинарне претраге. Ако постоји пар, одштампајте тај пар. иначе принт не постоји пар.
Испод је примена горњег приступа:
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)
Излаз
Pair elements are 11 and 4
Временска сложеност: О(н * пријава)
Помоћни простор: О(1)
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