# LeetCode – Range Addition (Java)

Assume you have an array of length n initialized with all 0's and are given k update operations.

Each operation is represented as a triplet: [startIndex, endIndex, inc] which increments each element of subarray A[startIndex ... endIndex] (startIndex and endIndex inclusive) with inc.

Return the modified array after all k operations were executed.

Java Solution 1 - Using a heap

```public int[] getModifiedArray(int length, int[][] updates) { int result[] = new int[length]; if(updates==null || updates.length==0) return result;   //sort updates by starting index Arrays.sort(updates, new Comparator<int[]>(){ public int compare(int[] a, int [] b){ return a[0]-b[0]; } });   ArrayList<int[]> list = new ArrayList<int[]>();   //create a heap sorted by ending index PriorityQueue<Integer> queue = new PriorityQueue<Integer>(new Comparator<Integer>(){ public int compare(Integer a, Integer b){ return updates[a][1]-updates[b][1]; } });   int sum=0; int j=0; for(int i=0; i<length; i++){ //substract value from sum when ending index is reached while(!queue.isEmpty() && updates[queue.peek()][1] < i){ int top = queue.poll(); sum -= updates[top][2]; }   //add value to sum when starting index is reached while(j<updates.length && updates[j][0] <= i){ sum = sum+updates[j][2]; queue.offer(j); j++; }   result[i]=sum; }   return result; }```

Time complexity is O(nlog(n)).

Java Solution 2

```public int[] getModifiedArray(int length, int[][] updates) { int[] result = new int[length]; if(updates==null||updates.length==0) return result;   for(int i=0; i<updates.length; i++){ result[updates[i][0]] += updates[i][2]; if(updates[i][1]<length-1){ result[updates[i][1]+1] -=updates[i][2]; } }   int v=0; for(int i=0; i<length; i++){ v += result[i]; result[i]=v; }   return result; }```

Time complexity is O(n).

Category >> Algorithms
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• Ankit Shah

why do we need heap here, can someone post the question with example ?

• Hayro

The second solution is super, very smart!