Write a javascript program for a given array arr[], count the number of pairs arr[i], arr[j] such that arr[i] + arr[j] is maximum and i < j.
Example:
Input : arr[] = {1, 1, 1, 2, 2, 2}
Output: 3
Explanation: The maximum possible pair sum where i<j is 4, which is given by 3 pairs, so the answer is 3 the pairs are (2, 2), (2, 2) and (2, 2)Input: arr[] = {1, 4, 3, 3, 5, 1}
Output: 1
Explanation: The pair 4, 5 yields the maximum sum i.e, 9 which is given by 1 pair only
Javascript Program for Number of pairs with maximum sum using Naive Approach:
Traverse a loop i from 0 to n, i.e length of the array and another loop j from i+1 to n to find all possible pairs with i<j. Find the pair with the maximum possible sum, again traverse for all pairs and keep the count of the number of pairs which gives the pair sum equal to maximum .
Below is the Implementation of the above Approach:
Javascript
// JavaScript program to count pairs// with maximum sum.// Function to find the number of// maximum pair sumsfunction sum(a, n){ // traverse through all the pairs let maxSum = Number.MIN_VALUE; for(let i = 0; i < n; i++) for(let j = i + 1; j < n; j++) maxSum = Math.max(maxSum, a[i] + a[j]); // Traverse through all pairs and // keep a count of the number of // maximum pairs let c = 0; for(let i = 0; i < n; i++) for(let j = i + 1; j < n; j++) if (a[i] + a[j] == maxSum) c++; return c;}// Driver Codelet array = [ 1, 1, 1, 2, 2, 2 ];let n = array.length;document.write(sum(array, n));// This code is contributed by code_hunt |
3
Time complexity: O(n2)
Auxiliary Space: O(1)
Efficient Method:
- Maximum element is always part of solution.
- If maximum element appears more than once, then result is maxCount * (maxCount – 1)/2. We basically need to choose 2 elements from maxCount (maxCountC2).
- If maximum element appears once, then result is equal to count of second maximum element. We can form a pair with every second max and max.
Below is the Implementation of the above Approach:
Javascript
// Javascript program to count pairs with maximum sum.// function to find the number// of maximum pair sumsfunction sum(a, n){ // Find maximum and second maximum // elements. Also find their counts. let maxVal = a[0], maxCount = 1; let secondMax = Number.MIN_VALUE; let secondMaxCount = 0; for (let i = 1; i < n; i++) { if (a[i] == maxVal) maxCount++; else if (a[i] > maxVal) { secondMax = maxVal; secondMaxCount = maxCount; maxVal = a[i]; maxCount = 1; } else if (a[i] == secondMax) { secondMax = a[i]; secondMaxCount++; } else if (a[i] > secondMax) { secondMax = a[i]; secondMaxCount = 1; } } // If maximum element appears // more than once. if (maxCount > 1) return maxCount * parseInt((maxCount - 1) / 2, 10); // If maximum element appears // only once. return secondMaxCount;}let array = [ 1, 1, 1, 2, 2, 2, 3 ];let n = array.length;document.write(sum(array, n));// This code is contributed by divyesh072019. |
3
Time complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Number of pairs with maximum sum for more details!
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