How can I find the minimum index of the array in this case?

Question

We are given an array with n values.

Example: [1,4,5,6,6]

For each index i of the array a ,we construct a new element of array b such that,

b[i]= [a[i]/1] + [a[i+1]/2] + [a[i+2]/3] + ⋯ + [a[n]/(n−i+1)] where [.] denotes the greatest integer function.

We are given an integer k as well.

We have to find the minimum i such that b[i] ≤ k.

I know the brute-force O(n^2) algorithm (to create the array - 'b'), can anybody suggest a better time complexity and way solve it?

For example, for the input [1,2,3],k=3, the output is 1(minimum-index).

Here, a[1]=1; a[2]=2; a[3]=3;

Now, b[1] = [a[1]/1] + [a[2]/2] + [a[3]/3] = [1/1] + [2/2] + [3/3] = 3;

b[2] = [a[2]/1] + [a[3]/2] = [2/1] + [3/2] = 3;

b[3] = [a[3]/1] = [3/1] = 3 (obvious)

Now, we have to find the index i such that b[i]<=k , k='3' , also b[1]<=3, henceforth, 1 is our answer! :-)

Constraints : - Time limits: -(2-seconds) , 1 <= a[i] <= 10^5, 1 <= n <= 10^5, 1 <= k <= 10^9

Answer 1

This probably cannot reach the efficiency of David Eisenstat's answer but since I spent quite a long time figuring out an implementation, I thought I'd leave it up anyway. As it is, it seems about O(n^2).

The elements of b[i] may be out of order, but sections of them are not:

[a[1]/1] + [a[2]/2] + [a[3]/3]
           |------ s2_1 -----|
                      |-s1_1-|

           [a[2]/1] + [a[3]/2]
           |------ s2_2 -----|
                      |-s1_2-|

                      [a[3]/1]
                      |-s1_3-|


s2_1 < s2_2

s1_1 < s1_2 < s1_3

Binary search for k on s1. Any result with an s1_i greater than k will rule out a section of ordered rows (rows are b_is).

Binary search for k on s2 on the remaining rows. Any result with an s2_i greater than k will rule out a section of ordered rows (rows are b_is).

This wouldn't help much since in the worst case, we'd have O(n^2 * log n) complexity, greater than O(n^2).

But we can also search horizontally. If we know that b_i ≤ k, then it will rule out both all rows with greater or equal length and the need to search smaller s(m)s, not because smaller s(m)s cannot produce a sum >= k, but because they will necessarily produce one with a higher i and we are looking for the minimum i.

JavaScript code:

var sum_width_iterations = 0
var total_width_summed = 0

var sum_width_cache = {}

function sum_width(A, i, width){
  let key = `${i},${width}`
  
  if (sum_width_cache.hasOwnProperty(key))
    return sum_width_cache[key]
    
  sum_width_iterations++
  total_width_summed += width

  let result = 0
  for (let j=A.length-width; j<A.length; j++)
    result += ~~(A[j] / (j + 1 - i))
  return sum_width_cache[key] = result
}

function get_b(A){
  let result = []
  A.map(function(a, i){
    result.push(sum_width(A, i, A.length - i))
})
  return result
}

function find_s_greater_than_k(A, width, low, high, k){
  let mid = low + ((high - low) >> 1)
  let s = sum_width(A, mid, width)
  while (low <= high){
    mid = low + ((high - low) >> 1)
    s = sum_width(A, mid, width)
    if (s > k)
      high = mid - 1
    else
      low = mid + 1
  }
  return [mid, s]
}

function f(A, k, l, r){
  let n = A.length
  
  if (l > r){
  
console.log(`l > r: l, r: ${l}, ${r}`)
  
    return [n + 1, Infinity]
  }
  
  let width = n - l
  
console.log(`\n(call) width, l, r: ${width}, ${l}, ${r}`)

  let mid = l + ((r - l) >> 1)
  let mid_width = n - mid
  
console.log(`mid: ${mid}`)
console.log('mid_width: ' + mid_width)

  let highest_i = n - mid_width
  let [i, s] = find_s_greater_than_k(A, mid_width, 0, highest_i, k)

console.log(`hi_i, s,i,k: ${highest_i}, ${s}, ${i}, ${k}`)

  if (mid_width == width)
    return [i, s]

  // either way we need to look left
  // and down

console.log(`calling left`)

  let [li, ls] = f(A, k, l, mid - 1)

  // if i is the highest, width is
  // the width of b_i

console.log(`got left: li, ls, i, high_i: ${li}, ${ls}, ${i}, ${highest_i}`)

  if (i == highest_i){

console.log(`i == highest_i, s <= k: ${s <= k}`)

    // b_i is small enough
    if (s <= k){
      if (ls <= k)
        return [li, ls]
      else
        return [i, s]

    // b_i is larger than k
    } else {

console.log(`b_i > k`)

      let [ri, rs] = f(A, k, mid + 1, r)

console.log(`ri, rs: ${ri}, ${rs}`)

      if (ls <= k)
        return [li, ls]
      else if (rs <= k)
        return [ri, rs]
      else
        return [i, s]
    }
    
  // i < highest_i
  } else {

  console.log(`i < highest_i: high_i, i, s, li, ls, mid, mid_width, width, l, r: ${highest_i}, ${i}, ${s}, ${li}, ${ls}, ${mid}, ${mid_width}, ${width}, ${l}, ${r}`)
  
    // get the full sum for this b
    let b_i = sum_width(A, i, n - i)
      
console.log(`b_i: ${b_i}`)
  
    // suffix sum is less than k
    // so we cannot rule out either side
    if (s < k){
    
console.log(`s < k`)

      let ll = l
      let lr = mid - 1
      let [lli, lls] = f(A, k, ll, lr)
      
console.log(`ll, lr, lli, lls: ${ll}, ${lr}, ${lli}, ${lls}`)
      
      // b_i is a match so we don't
      // need to look to the right
      if (b_i <= k){
      
console.log(`b_i <= k: i, b_i: ${i}, ${b_i}`)
      
        if (lls <= k)
          return [lli, lls]
        else
          return [i, b_i]
      
      // b_i > k
      } else {
      
console.log(`b_i > k: i, b_i: ${i}, ${b_i}`)
      
        let rl = mid + 1
        let rr = r
        let [rri, rrs] = f(A, k, rl, rr)
      
console.log(`rl, rr, rri, rrs: ${rl}, ${rr}, ${rri}, ${rrs}`)
      
        // return the best of right
        // and left sections
        if (lls <= k)
          return [lli, lls]
        else if (rrs <= k)
          return [rri, rrs]
        else
          return [i, b_i]
      }
    
    // suffix sum is greater than or
    // equal to k so we can rule out
    // this and all higher rows (`b`s)
    // that share this suffix
    } else {
    
console.log(`s >= k`)

      let ll = l
      // the suffix rules out b_i
      // and above
      let lr = i - 1
      let [lli, lls] = f(A, k, ll, lr)
      
console.log(`ll, lr, lli, lls: ${ll}, ${lr}, ${lli}, ${lls}`)

      let rl = highest_i + 1
      let rr = r
      let [rri, rrs] = f(A, k, rl, rr)
      
console.log(`rl, rr, rri, rrs: ${rl}, ${rr}, ${rri}, ${rrs}`)

      // return the best of right
      // and left sections
      if (lls <= k)
        return [lli, lls]
      else if (rrs <= k)
        return [rri, rrs]
      else
        return [i, b_i]
    }
  }
}

let lst = [1, 2, 3, 1]
     // b [3, 3, 3, 1]
     
lst = [ 1, 34,  3,  2,  9, 21, 3, 2, 2, 1]
//  b [23, 41, 12, 13, 20, 22, 4, 3, 2, 1]

console.log(
  JSON.stringify(f(lst, 20, 0, lst.length)))

console.log(`sum_width_iterations: ${sum_width_iterations}`)

console.log(`total_width_summed: ${total_width_summed}`)

Answer 2

Here's an O(n √A)-time algorithm to compute the b array where n is the number of elements in the a array and A is the maximum element of the a array.

This algorithm computes the difference sequence of the b array (∆b = b[0], b[1] - b[0], b[2] - b[1], ..., b[n-1] - b[n-2]) and derives b itself as the cumulative sums. Since the differences are linear, we can start with ∆b = 0, 0, ..., 0, loop over each element a[i], and add the difference sequence for [a[i]], [a[i]/2], [a[i]/3], ... at the appropriate spot. The key is that this difference sequence is sparse (less than 2√a[i] elements). For example, for a[i] = 36,

>>> [36//j for j in range(1,37)]
[36, 18, 12, 9, 7, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
>>> list(map(operator.sub,_,[0]+_[:-1]))
[36, -18, -6, -3, -2, -1, -1, -1, 0, -1, 0, 0, -1, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

We can derive the difference sequence from a subroutine that, given a positive integer r, returns all maximal pairs of positive integers (p, q) such that pq ≤ r.

See complete Python code below.

def maximal_pairs(r):
    p = 1
    q = r
    while p < q:
        yield (p, q)
        p += 1
        q = r // p
    while q > 0:
        p = r // q
        yield (p, q)
        q -= 1


def compute_b_fast(a):
    n = len(a)
    delta_b = [0] * n
    for i, ai in enumerate(a):
        previous_j = i
        for p, q in maximal_pairs(ai):
            delta_b[previous_j] += q
            j = i + p
            if j >= n:
                break
            delta_b[j] -= q
            previous_j = j
    for i in range(1, n):
        delta_b[i] += delta_b[i - 1]
    return delta_b


def compute_b_slow(a):
    n = len(a)
    b = [0] * n
    for i, ai in enumerate(a):
        for j in range(n - i):
            b[i + j] += ai // (j + 1)
    return b


for n in range(1, 100):
    print(list(maximal_pairs(n)))

lst = [1, 34, 3, 2, 9, 21, 3, 2, 2, 1]
print(compute_b_fast(lst))
print(compute_b_slow(lst))

Answer 3

Why should calculating b[i] lead to O(n²)? If i = 1, it takes n steps. If i = n, it takes one step to calculate b[i]...

You could improve your calculation when you abort the sum on the condition Sum > k.

Let a in N^n
Let k in N

for (i1 := 1; i1 <= n; i1++)
  b := 0
  for (i2 :=i1; i2 <= n; i2++)   // This loop is the calculation of b[i]
    b := b + ceil(a[i2]/(i2 + 1))
    if (b > k)
      break
  if (i2 == n)
    return i1