Algorithm
Problem Name: 220. Contains Duplicate III
You are given an integer array nums and two integers indexDiff and valueDiff.
Find a pair of indices (i, j) such that:
i != j,abs(i - j) <= indexDiff.abs(nums[i] - nums[j]) <= valueDiff, and
Return true if such pair exists or false otherwise.
Example 1:
Input: nums = [1,2,3,1], indexDiff = 3, valueDiff = 0 Output: true Explanation: We can choose (i, j) = (0, 3). We satisfy the three conditions: i != j --> 0 != 3 abs(i - j) <= indexDiff --> abs(0 - 3) <= 3 abs(nums[i] - nums[j]) <= valueDiff --> abs(1 - 1) <= 0
Example 2:
Input: nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3 Output: false Explanation: After trying all the possible pairs (i, j), we cannot satisfy the three conditions, so we return false.
Constraints:
2 <= nums.length <= 105-109 <= nums[i] <= 1091 <= indexDiff <= nums.length0 <= valueDiff <= 109
Code Examples
#1 Code Example with C Programming
Code -
C Programming
typedef struct {
int val;
int idx;
} e_t;
int cmp(void const *a, void const *b) {
return ((const e_t *)a)->val < ((const e_t *)b)->val ? -1 :
((const e_t *)a)->val > ((const e_t *)b)->val ? 1 :
((const e_t *)a)->idx < ((const e_t *)b)->idx ? -1 : 1;
}
bool containsNearbyAlmostDuplicate(int* nums, int numsSize, int k, int t) {
e_t *p;
int i, j, d, f;
p = malloc(numsSize * sizeof(e_t));
//assert(p);
for (i = 0; i < numsSize; i ++) {
p[i].val = nums[i];
p[i].idx = i;
}
qsort(p, numsSize, sizeof(e_t), cmp);
f = 0;
for (i = 0; !f && i < numsSize - 1; i ++) {
//printf("\n%d: %d\n", p[i].idx, p[i].val);
for (j = i + 1;
!f &&
j < numsSize &&
((p[i].val > 0 && p[j].val - p[i].val <= t) ||
p[j].val <= t + p[i].val);
j ++) {
//printf("%d: %d\n", p[j].idx, p[j].val);
d = p[j].idx - p[i].idx;
if (d < 0) d = -d;
if (d <= k) f = 1;
}
}
free(p);
return f;
}
Input
nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
Output
true
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const containsNearbyAlmostDuplicate = function(nums, k, t) {
if (k < 1 || t < 0) {
return false
}
const array = new Map()
const num = 10 ** 10
for (let i = 0, iL = nums.length; i < iL; ++i) {
const noNegative = nums[i] + num
const factor = Math.floor(noNegative / (t + 1))
if (
array.has(factor) ||
(array.has(factor - 1) && noNegative - array.get(factor - 1) <= t) ||
(array.has(factor + 1) && array.get(factor + 1) - noNegative <= t)
) {
return true
}
if (array.size >= k) {
array.delete(Math.floor((nums[i - k] + num) / (t + 1)))
}
array.set(factor, noNegative)
}
return false
}
Input
nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
Output
true
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def containsNearbyAlmostDuplicate(self, nums, k, t):
if t < 0: return False
d = {}
for i in range(len(nums)):
m = nums[i] // (t + 1)
if m in d or (m - 1 in d and nums[i] - d[m - 1] <= t) or (m + 1 in d and d[m + 1] - nums[i] <= t):
return True
d[m] = nums[i]
if i >= k: del d[nums[i - k] // (t + 1)]
return False
Input
nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
Output
false
#4 Code Example with C# Programming
Code -
C# Programming
using System;
using System.Linq;
namespace LeetCode
{
public class _0220_ContainsDuplicateIII
{
public bool ContainsNearbyAlmostDuplicate(int[] nums, int k, int t)
{
var arr = nums.Select((num, index) => new { num, index })
.OrderBy(u => u.num)
.ToArray();
for (int i = 0; i < nums.Length; i++)
for (int j = i + 1; j < nums.Length && arr[j].num - (long)arr[i].num <= t; j++)
if (Math.Abs(arr[i].index - arr[j].index) <= k)
return true;
return false;
}
}
}
Input
nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
Output
false