Algorithm to calculate number of intersecting discs

Question

Given an array A of N integers we draw N discs in a 2D plane, such that i-th disc has center in (0,i) and a radius

Answer 1

Probably extremely fast. O(N). But you need to check it out. 100% on Codility. Main idea: 1. At any point of the table, there are number of circles "opened" till the right edge of the circle, lets say "o". 2. So there are (o-1-used) possible pairs for the circle in that point. "used" means circle that have been processed and pairs for them counted.

  public int solution(int[] A) { 
    final int N = A.length;
    final int M = N + 2;
    int[] left  = new int[M]; // values of nb of "left"  edges of the circles in that point
    int[] sleft = new int[M]; // prefix sum of left[]
    int il, ir;               // index of the "left" and of the "right" edge of the circle

    for (int i = 0; i < N; i++) { // counting left edges
      il = tl(i, A);
      left[il]++;
    }

    sleft[0] = left[0];
    for (int i = 1; i < M; i++) {// counting prefix sums for future use
      sleft[i]=sleft[i-1]+left[i];
    }
    int o, pairs, total_p = 0, total_used=0;
    for (int i = 0; i < N; i++) { // counting pairs
      ir = tr(i, A, M);
      o  = sleft[ir];                // nb of open till right edge
      pairs  = o -1 - total_used;
      total_used++;
      total_p += pairs;
    }
    if(total_p > 10000000){
      total_p = -1;
    }
    return total_p;
  }

    private int tl(int i, int[] A){
    int tl = i - A[i]; // index of "begin" of the circle
      if (tl < 0) {
        tl = 0;
      } else {
        tl = i - A[i] + 1;
      }
    return tl;
  }
  int tr(int i, int[] A, int M){
    int tr;           // index of "end" of the circle
      if (Integer.MAX_VALUE - i < A[i] || i + A[i] >= M - 1) {
        tr = M - 1;
      } else {
        tr = i + A[i] + 1;
      }
      return tr;
  }