Count good triplets
Question
Given an array of integers arr, and three integers a, b and c. You need to find the number of good triplets.
A triplet (arr[i], arr[j], arr[k]) is good if the following conditions are true:
0 <= i < j < k < arr.length|arr[i] - arr[j]| <= a|arr[j] - arr[k]| <= b|arr[i] - arr[k]| <= c
Where |x| denotes the absolute value of x.
Return the number of good triplets.
Example 1:
Input: arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3 Output: 4 Explanation: There are 4 good triplets: [(3,0,1), (3,0,1), (3,1,1), (0,1,1)].
Example 2:
Input: arr = [1,1,2,2,3], a = 0, b = 0, c = 1 Output: 0 Explanation: No triplet satisfies all conditions.
Constraints:
3 <= arr.length <= 1000 <= arr[i] <= 10000 <= a, b, c <= 1000
Analysis
Approach 1: Enumeration
Intuition
Using O(n3) loops to enumerate all (i,j,k) in sequence, where 0≤i<j<k<arr.length, for each set of (i,j,k), determine whether arr[i], arr[j], and arr[k] satisfy the condition.
Finally, calculate the total number of all triplets that meet the conditions.
class Solution:
def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
n = len(arr)
countGoodTriplets = 0
for i in range(n):
for j in range(i + 1, n):
if abs(arr[i] - arr[j]) <= a:
for k in range(j + 1, n):
if abs(arr[j] - arr[k]) <= b and abs(arr[i] - arr[k]) <= c:
countGoodTriplets += 1
return countGoodTripletsApproach 2: Optimized enumeration
We need to eliminate one loop or make one of them constant-time.
We could try fixing (j,k) and then counting valid i < j. That \(O(n^2)\) pairs, and if we can count the \(i\)s in \(O(1)\) each, we down to \(O(n^2)\) total.
Translate the \(i\)-constraints into an interval
For a given \((j,k)\), the condition on \(i\) are \[|arr[i] - arr[j]| \leq a\] so, \[arr[i] \in [arr[j]-a, arr[j]+a]\]
and \[|arr[i]-arr[k]|\leq c\] so, \[arr[i]\in[arr[k]-c, arr[k]+c]\]
Their intersection is a single interval \([l, r]\). So we just need to count how many prior \(i<j\) have \(arr[i] \in [l, r]\).
Choose a data structure for last range-count
Since arr[i] is small-range, maintain a frequency array freq[0..M] for all seen indices \(<j\), and its prefix sums sum[v] =\(\Sigma_{u=0}^v freq[u]\)
Then the count in \([l, r]\) is simply sum[r]-sum[l-1] in \(O(1)\).
Ensure the \(i<j\) ordering
We iterate \(j\) from 1 to \(n\). Before handling any \((j, k)\) pairs, our freq/sum reflects exact indices 1 through \(j-1\).
For each \(k>j\), check \(|arr[j] - arr[k]|\leq b\). If it passes, compute \([l, r]\) and do the \(O(1)\) range-sum query.
After finishing all \(k\) for this \(j\), we insert \(arr[j]\) into freq (and update sum), before moving onto \(j + 1\).
class Solution:
def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
# ans: total count of valid (i,j,k)
ans = 0
n = len(arr)
# total[v] will hold the prefix-sum of frequencies:
# total[v] = number of i<j with arr[i] ≤ v
# we assume arr[i] ∈ [0..1000], so we size total to 1001
total = [0] * 1001
# Move j from 0 to n-1
for j in range(n):
# Pair j with every k>j
for k in range(j + 1, n):
# First check the |arr[j] - arr[k]| ≤ b constraint
if abs(arr[j] - arr[k]) <= b:
# Compute the interval of allowed arr[i] from j’s constraint
lj, rj = arr[j] - a, arr[j] + a
# Compute the interval of allowed arr[i] from k’s constraint
lk, rk = arr[k] - c, arr[k] + c
# Intersection [l..r] of the two intervals
# also clamp to [0..1000] to stay in array bounds
l = max(0, lj, lk)
r = min(1000, rj, rk)
# If the intersection is non-empty, count how many
# prior i<j have arr[i] in [l..r] via prefix sums
if l <= r:
if l == 0:
ans += total[r]
else:
ans += total[r] - total[l - 1]
# After processing all k for this j, we “add” arr[j] into our
# prefix-sum structure so that future iterations see it.
# We do this by incrementing total[v] for all v ≥ arr[j].
# That way total[v] remains = # of arr[i] ≤ v for i<next j.
for v in range(arr[j], 1001):
total[v] += 1
return ans