У нас є масив цілих чисел, і ми повинні знайти два таких елементи в масиві, щоб сума цих двох елементів дорівнювала сумі решти елементів у масиві.
приклади:
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 просте рішення полягає в тому, щоб розглянути кожну пару по одному, знайти її суму та порівняти суму з сумою решти елементів. Якщо ми знаходимо пару, сума якої дорівнює решті елементів, ми друкуємо пару та повертаємо true. Часова складність цього рішення дорівнює O(n3)
Ан ефективне рішення це знайти суму всіх елементів масиву. Нехай ця сума буде «сумою». Тепер завдання зводиться до пошуку пари, сума якої дорівнює sum/2.
Інша оптимізація полягає в тому, що пара може існувати, лише якщо сума всього масиву парна, оскільки ми фактично ділимо його на дві частини з рівною сумою.
- Знайдіть суму всього масиву. Нехай ця сума буде "сумою"
- Якщо сума непарна, поверніть false.
- Знайдіть пару із сумою, що дорівнює «сумі/2», використовуючи розглянутий метод на основі хешування тут як метод 2. Якщо знайдено пару, вивести її та повернути true.
- Якщо пари не існує, повертає false.
Нижче наведено реалізацію вищевказаних кроків.
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>
Вихід
Pair elements are 4 and 11
Часова складність: O(n) . невпорядкований_набір реалізовано за допомогою хешування. Тут припускається, що хеш-пошук і вставка складності за часом є O(1).
Допоміжний простір: O(n)
Інший ефективний підхід (оптимізація простору): Спочатку ми відсортуємо масив для Двійковий пошук . Потім ми повторимо весь масив і перевіримо, чи існує в масиві індекс, який поєднується з i таким чином, що arr[index] + a[i] == Решта сума масиву. Ми можемо використовувати двійковий пошук, щоб знайти індекс у масиві, змінивши програму бінарного пошуку. Якщо пара існує, надрукуйте цю пару. інакше print не існує пари.
Нижче наведено реалізацію вищезазначеного підходу:
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)
Вихід
Pair elements are 11 and 4
Часова складність: O(n * logn)
Допоміжний простір: О(1)
код abs c
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