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authorJack Sangdahl <[email protected]>2025-02-08 18:00:46 -0700
committerJack Sangdahl <[email protected]>2025-02-08 18:00:46 -0700
commit38628b7a294223f7e1caa129f55753183a536fa6 (patch)
tree5939376ce83a3558c0b76bc8c995f7223809edc5
parentf72076c568e27c480a1154b527433077e3884aad (diff)
downloadleetcode-38628b7a294223f7e1caa129f55753183a536fa6.tar.gz
leetcode-38628b7a294223f7e1caa129f55753183a536fa6.zip
2364. count number of bad pairs
-rw-r--r--num_bad_pairs.c41
-rw-r--r--num_bad_pairs.cpp19
-rw-r--r--num_bad_pairs.py27
-rw-r--r--num_bad_pairs.rs15
4 files changed, 86 insertions, 16 deletions
diff --git a/num_bad_pairs.c b/num_bad_pairs.c
new file mode 100644
index 0000000..45292b2
--- /dev/null
+++ b/num_bad_pairs.c
@@ -0,0 +1,41 @@
+// 2364. Count Number of Bad Pairs
+#include "leetcode.h"
+
+/*
+ * you are given a 0-indexed integer array 'nums'. a pair of indices '(i, j)' is
+ * a bad pair if 'i < j' and 'j - i != nums[j] - nums[i]'. return the total
+ * number of bad pairs in 'nums'.
+ */
+
+struct hash {
+ int key;
+ int count;
+ UT_hash_handle hh;
+};
+
+long long countBadPairs(int *nums, int numsSize) {
+ struct hash *total = NULL;
+ long long ans = 0;
+ for (int i = numsSize; i >= 0; i--) {
+ int target = nums[i] - i;
+ struct hash *tmp = NULL;
+ HASH_FIND_INT(total, &target, tmp);
+ if (!tmp) {
+ tmp = (struct hash *)malloc(sizeof(struct hash));
+ tmp->key = target;
+ tmp->count = 1;
+ HASH_ADD_INT(total, key, tmp);
+ ans += numsSize - i - 1;
+ } else {
+ ans = numsSize - i - tmp->count - 1;
+ tmp->count++;
+ }
+ }
+ return ans;
+}
+
+int main() {
+ int n1[] = {4, 1, 3, 3}, n2[] = {1, 2, 3, 4, 5};
+ printf("%lld\n", countBadPairs(n1, ARRAY_SIZE(n1))); // expect: 5
+ printf("%lld\n", countBadPairs(n2, ARRAY_SIZE(n2))); // expect: 0
+}
diff --git a/num_bad_pairs.cpp b/num_bad_pairs.cpp
index 75ea430..144e6ef 100644
--- a/num_bad_pairs.cpp
+++ b/num_bad_pairs.cpp
@@ -1,19 +1,20 @@
+// 2364. Count Number of Bad Pairs
#include "leetcode.h"
/*
- * given: 0-indexed integer array
- * bad pair: (i, j) if
- * i < j && j - i != nums[j] - nums[i]
+ * you are given a 0-indexed integer array 'nums'. a pair of indices '(i, j)' is
+ * a bad pair if 'i < j' and 'j - i != nums[j] - nums[i]'. return the total
+ * number of bad pairs in 'nums'.
*/
class Solution {
public:
long long countBadPairs(vector<int> &nums) {
long long ans = 0;
- unordered_map<int, int> um;
+ unordered_map<int, int> mp;
for (int i = 0; i < nums.size(); ++i) {
- ans += (long long)i - um[nums[i] - i];
- ++um[nums[i] - i];
+ ans += i - mp[nums[i] - i];
+ mp[nums[i] - i]++;
}
return ans;
}
@@ -21,7 +22,7 @@ public:
int main() {
Solution obj;
- vector<int> nums1 = {4, 1, 3, 3}, nums2 = {1, 2, 3, 4, 5};
- cout << obj.countBadPairs(nums1) << endl; // expect: 5
- cout << obj.countBadPairs(nums2) << endl; // expect: 0
+ vector<int> n1 = {4, 1, 3, 3}, n2 = {1, 2, 3, 4, 5};
+ printf("%lld\n", obj.countBadPairs(n1)); // expect: 5
+ printf("%lld\n", obj.countBadPairs(n2)); // expect: 0
}
diff --git a/num_bad_pairs.py b/num_bad_pairs.py
new file mode 100644
index 0000000..6f3ae05
--- /dev/null
+++ b/num_bad_pairs.py
@@ -0,0 +1,27 @@
+# 2364. Count Number of Bad Pairs
+from collections import defaultdict
+
+"""
+you are given a 0-indexed integer array 'nums'. a pair of indices '(i, j)' is
+a bad pair if 'i < j' and 'j - i != nums[j] - nums[i]'. return the total
+number of bad pairs in 'nums'.
+"""
+
+
+class Solution(object):
+ def countBadPairs(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: int
+ """
+ mp, ans = defaultdict(int), 0
+ for i in range(len(nums)):
+ ans += i - mp[nums[i] - i]
+ mp[nums[i] - i] += 1
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.countBadPairs(nums=[4, 1, 3, 3]))
+ print(obj.countBadPairs(nums=[1, 2, 3, 4, 5]))
diff --git a/num_bad_pairs.rs b/num_bad_pairs.rs
index 8e1e7dd..88119d8 100644
--- a/num_bad_pairs.rs
+++ b/num_bad_pairs.rs
@@ -1,10 +1,11 @@
+// 2364. Count Number of Bad Pairs
use std::collections::HashMap;
struct Solution;
/*
- * given: 0-indexed integer array
- * bad pair: (i, j) if
- * i < j && j - i != nums[j] - nums[i]
+ * you are given a 0-indexed integer array 'nums'. a pair of indices '(i, j)' is
+ * a bad pair if 'i < j' and 'j - i != nums[j] - nums[i]'. return the total
+ * number of bad pairs in 'nums'.
*/
impl Solution {
@@ -22,8 +23,8 @@ impl Solution {
}
fn main() {
- let nums1 = vec![4,1,3,3];
- let nums2 = vec![1,2,3,4,5];
- println!("{}", Solution::count_bad_pairs(nums1)); //expect: 5
- println!("{}", Solution::count_bad_pairs(nums2)); //expect: 0
+ let n1 = vec![4, 1, 3, 3];
+ let n2 = vec![1, 2, 3, 4, 5];
+ println!("{}", Solution::count_bad_pairs(n1)); //expect: 5
+ println!("{}", Solution::count_bad_pairs(n2)); //expect: 0
}