Length of minimum subsequence that spans entire length of array?

Question

I'm given an integer array of length N, with elements representing the spans it could cover from 0 to N-1. so for an element a[i] the element spans a range from max(i-a[i],0) to min(i+a[i],N-1).

How to find the length of smallest subsequence that spans entire space ie from 0 to N-1;

Example : For this array [1,1,1,3,2,1,1,4] answer should be 2

This is what i've got so far, It doesn't work for all cases

int arr[] = {2,2,1,3,2,1,1,4,1,1,1,1,1,1,1,1,1,1}; // Fails for this case
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        int[] sortedIndices = IntStream.range(0, arr.length)
                .boxed().sorted((i, j) -> span(arr[j],j,arr.length) - span(arr[i],i,arr.length))
                .mapToInt(ele -> ele).toArray();

        int i=0;
        int ans = 0;
        while (min>0 || max<arr.length-1) {
            int ind = sortedIndices[i++];
            int val = arr[ind];

            int tmin = Math.max(0,ind-val);
            int tmax = Math.min(arr.length - 1, ind+val);
            if(tmin < min || tmax > max)
                ans++;

            min = Math.min(min, tmin);
            max = Math.max(max, tmax);
        }

        System.out.println(ans);

How did you get answer 2 for the first example? I see 4 (index range 3..6) — MBo, Apr 25 '19 at 07:09
You'll need at least two elements to cover entire span, like element 3 and 4 positioned in array at 3 & 7 respectively. — Mega Mbo, Apr 25 '19 at 08:03
So you really need not subarray (continuous) but subsequence. So this is [set cover problem](https://en.wikipedia.org/wiki/Set_cover_problem) — MBo, Apr 25 '19 at 08:07
Thanks for correcting ! Will edit for more clarity. No idea about set cover problem, will look into it — Mega Mbo, Apr 25 '19 at 08:12
Possible duplicate of [Finding the minimal coverage of an interval with subintervals](https://stackoverflow.com/questions/293168/finding-the-minimal-coverage-of-an-interval-with-subintervals) — גלעד ברקן, Apr 25 '19 at 15:06
Also related https://cs.stackexchange.com/questions/9531/finding-the-minimum-subset-of-intervals-covering-the-whole-set — גלעד ברקן, Apr 25 '19 at 15:07

MBo · Accepted Answer · 2019-04-25T09:43:10.893

Seems that problem might solved with simple greedy algorithm.

For every array entry create interval with start and finish fields (a[i].start = i-arr[i] etc)

Sort intervals by start field.

Find interval with the most right finish among those covering 0, make maxx=a[i].finish.

Scan intervals with start in 0..maxx range, choose one with the rightmost finish.

Continue.

Quick-made Python implementation (perhaps some more checks are needed)

ar = [2,2,1,3,2,1,4,4]
n = len(ar)
a = [(max(0, i-ar[i]), min(i+ar[i], n-1)) for i in range(n)]
print(ar)
a.sort()
print(a)
cnt = 0
maxx = -1
left = 0
i = 0
while i < n and maxx < n - 1:
    while i < n and a[i][0] <= left and maxx < n - 1:
        maxx = max(a[i][1], maxx)
        i+=1
    print("maxx ", maxx)
    left = maxx
    cnt += 1

print("cover ", cnt)

[2, 2, 1, 3, 2, 1, 4, 4]
[(0, 2), (0, 3), (0, 6), (1, 3), (2, 6), (2, 7), (3, 7), (4, 6)]
maxx  6
maxx  7
cover  2

Amongst items covering 0 the rightmost border 6 is for arr[3]. Amongst items covering 6 (starting from 1 to 6) the rightmost border is for arr[6] or arr[7], so count is 2 — MBo, Apr 25 '19 at 09:22

Length of minimum subsequence that spans entire length of array?

1 Answers1