给定一个未排序的整数数组。任务是找到总和相等的任意两个不重叠的对。
如果两对中的所有元素都位于不同的索引处，则说两对是不重叠的。也就是说，如果i≠j≠m≠n，则说成对(A i ，A j )和对(A m ，A n )是不重叠的。
例子：
Input: arr[] = {8, 4, 7, 8, 4}
Output: Pair First(4, 8)
Pair Second (8, 4)
Input: arr[] = {8, 4, 7, 4}
Output: No such non-overlapping pairs present.
Note: (8, 4) and (8, 4) forms overlapping pair
as index of 8 is same in both pairs.
想法是使用键值对映射存储所有出现的和。映射中的键将存储总和，并且对应的值将是具有该总和的索引对(i，j)的列表。
- 这个想法是遍历数组并生成所有可能的对。
- 如果找到了新的和，则将其直接插入到地图中，否则请检查先前遇到的任何一对与该和是否与当前对不重叠。
- 如果不是，则打印两对。
下面是上述方法的实现：
C++
// C++ programs to find two non-overlapping
// pairs having equal sum in an Array
#include
#include
#include
using namespace std;
typedef pair Pair;
// Function to find two non-overlapping
// with same sum in an array
void findPairs(int arr[], int n)
{
// first create an empty map
// key -> which is sum of a pair of
// elements in the array
// value -> vector storing index of
// every pair having that sum
unordered_map > map;
// consider every pair (arr[i], arr[j])
// and where (j > i)
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// calculate sum of current pair
int sum = arr[i] + arr[j];
// if sum is already present in the map
if (map.find(sum) != map.end()) {
// check every pair having equal sum
for (auto pair : map.find(sum)->second) {
int m = pair.first, n = pair.second;
// if pairs don't overlap,
// print them and return
if ((m != i && m != j) && (n != i && n != j)) {
cout << "Pair First(" << arr[i] << ", "
<< arr[j] << ")\nPair Second ("
<< arr[m] << ", " << arr[n] << ")";
return;
}
}
}
// Insert current pair into the map
map[sum].push_back({ i, j });
}
}
// If no such pair found
cout << "No such non-overlapping pairs present";
}
// Driver Code
int main()
{
int arr[] = { 8, 4, 7, 8, 4 };
int size = sizeof(arr) / sizeof(arr[0]);
findPairs(arr, size);
return 0;
} Java
// Java program to find two non-overlapping
// pairs having equal sum in an Array
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
// Declare a pair
class Pair {
public int x, y;
Pair(int x, int y)
{
this.x = x;
this.y = y;
}
}
class GFG {
// Function to find two non-overlapping
// pairs with same sum in an array
public static void findPairs(int[] A)
{
// first create an empty map
// key -> which is sum of a pair
// of elements in the array
// value -> list storing index of
// every pair having that sum
Map > map = new HashMap<>();
// consider every pair (A[i], A[j]) where (j > i)
for (int i = 0; i < A.length - 1; i++) {
for (int j = i + 1; j < A.length; j++) {
// calculate sum of current pair
int sum = A[i] + A[j];
// if sum is already present in the map
if (map.containsKey(sum)) {
// check every pair having desired sum
for (Pair pair : map.get(sum)) {
int x = pair.x;
int y = pair.y;
// if pairs don't overlap, print
// them and return
if ((x != i && x != j) && (y != i && y != j)) {
System.out.println("Pair First(" + A[i] + ", "
+ A[j] + ")");
System.out.println("Pair Second (" + A[x] + ", "
+ A[y] + ")");
return;
}
}
}
// Insert current pair into the map
map.putIfAbsent(sum, new ArrayList<>());
map.get(sum).add(new Pair(i, j));
}
}
System.out.print("No such non-overlapping pairs present");
}
// Driver Code
public static void main(String[] args)
{
int[] A = { 8, 4, 7, 8, 4 };
findPairs(A);
}
} Python3
# Python3 programs to find two non-overlapping
# pairs having equal Sum in an Array
# Function to find two non-overlapping
# with same Sum in an array
def findPairs(arr, size):
# first create an empty Map
# key -> which is Sum of a pair of
# elements in the array
# value -> vector storing index of
# every pair having that Sum
Map = {}
# consider every pair (arr[i], arr[j])
# and where (j > i)
for i in range(0, size - 1):
for j in range(i + 1, size):
# calculate Sum of current pair
Sum = arr[i] + arr[j]
# if Sum is already present in the Map
if Sum in Map:
# check every pair having equal Sum
for pair in Map[Sum]:
m, n = pair
# if pairs don't overlap,
# print them and return
if ((m != i and m != j) and
(n != i and n != j)):
print("Pair First ({}, {})".
format(arr[i], arr[j]))
print("Pair Second ({}, {})".
format(arr[m], arr[n]))
return
# Insert current pair into the Map
if Sum not in Map:
Map[Sum] = []
Map[Sum].append((i, j))
# If no such pair found
print("No such non-overlapping pairs present")
# Driver Code
if __name__ == "__main__":
arr = [8, 4, 7, 8, 4]
size = len(arr)
findPairs(arr, size)
# This code is contributed by Rituraj JainC#
// C# program to find two non-overlapping
// pairs having equal sum in an Array
using System;
using System.Collections.Generic;
// Declare a pair
class Pair
{
public int x, y;
public Pair(int x, int y)
{
this.x = x;
this.y = y;
}
}
class GFG{
// Function to find two non-overlapping
// pairs with same sum in an array
public static void findPairs(int[] A)
{
// First create an empty map
// key -> which is sum of a pair
// of elements in the array
// value -> list storing index of
// every pair having that sum
Dictionary> map = new Dictionary>();
// Consider every pair (A[i], A[j])
// where (j > i)
for(int i = 0; i < A.Length - 1; i++)
{
for(int j = i + 1; j < A.Length; j++)
{
// Calculate sum of current pair
int sum = A[i] + A[j];
// If sum is already present in the map
if (map.ContainsKey(sum))
{
// Check every pair having desired sum
foreach(Pair pair in map[sum])
{
int x = pair.x;
int y = pair.y;
// If pairs don't overlap, print
// them and return
if ((x != i && x != j) &&
(y != i && y != j))
{
Console.WriteLine("Pair First(" + A[i] +
", " + A[j] + ")");
Console.WriteLine("Pair Second (" + A[x] +
", " + A[y] + ")");
return;
}
}
}
// Insert current pair into the map
map[sum] = new List();
map[sum].Add(new Pair(i, j));
}
}
Console.Write("No such non-overlapping pairs present");
}
// Driver Code
public static void Main(String[] args)
{
int[] A = { 8, 4, 7, 8, 4 };
findPairs(A);
}
}
// This code is contributed by Princi Singh 输出：
Pair First(4, 8)
Pair Second (8, 4)