Given an unsorted array, find the largest pair sum.
Solution 1. O(n*logn) runtime, using sorting
Solution 2. O(n) runtime, using heapify (max priority queue)
Option 1. Use Java priority queue API, easy to implement but uses O(n) extra memory.
Option 2. Since we are given an array, heapify the array to a max pq. This does not use extra memory.
Heapify takes O(n) runtime;
delete the max value from the heapified array takes O(logn) runtime as we need to percolate down the
new head element.
1 public class Solution { 2 public int findLargestPair(int[] A){ 3 if(A == null || A.length <= 1){ 4 return Integer.MIN_VALUE; 5 } 6 heapify(A); 7 int firstMax = A[0], secondMax = 0; 8 A[0] = A[A.length - 1]; 9 percolateDown(A, A.length - 1, 0); 10 secondMax = A[0]; 11 return firstMax + secondMax; 12 } 13 private void heapify(int[] A){ 14 for(int i = A.length / 2 - 1; i >= 0; i--){ 15 percolateDown(A, A.length, i); 16 } 17 } 18 private void percolateDown(int[] A, int len, int idx){ 19 int maxIdx = idx; 20 int leftChildIdx = 2 * idx + 1; 21 int rightChildIdx = 2 * idx + 2; 22 if(leftChildIdx < len && A[leftChildIdx] > A[maxIdx]){ 23 maxIdx = leftChildIdx; 24 } 25 if(rightChildIdx < len && A[rightChildIdx] > A[maxIdx]){ 26 maxIdx = rightChildIdx; 27 } 28 if(maxIdx != idx){ 29 int temp = A[idx]; 30 A[idx] = A[maxIdx]; 31 A[maxIdx] = temp; 32 percolateDown(A, len, maxIdx); 33 } 34 } 35 }
Solution 3. O(n) runtime, O(1) space, scan the input array once and upate the largest and second largest values along the way.
1 public int findLargestPair(int[] A){ 2 if(A == null || A.length <= 1){ 3 return Integer.MIN_VALUE; 4 } 5 int firstMax = Math.max(A[0], A[1]); 6 int secondMax = Math.min(A[0], A[1]); 7 for(int i = 2; i < A.length; i++){ 8 if(A[i] >= firstMax){ 9 secondMax = firstMax; 10 firstMax = A[i]; 11 } 12 else if(A[i] > secondMax){ 13 secondMax = A[i]; 14 } 15 } 16 return firstMax + secondMax; 17 }