Algorithm

Problem Name: 646. Maximum Length of Pair Chain

You are given an array of `n` pairs `pairs` where `pairs[i] = [lefti, righti]` and `lefti < righti`.

A pair `p2 = [c, d]` follows a pair `p1 = [a, b]` if `b < c`. A chain of pairs can be formed in this fashion.

Return the length longest chain which can be formed.

You do not need to use up all the given intervals. You can select pairs in any order.

Example 1:

```Input: pairs = [[1,2],[2,3],[3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4].
```

Example 2:

```Input: pairs = [[1,2],[7,8],[4,5]]
Output: 3
Explanation: The longest chain is [1,2] -> [4,5] -> [7,8].
```

Constraints:

• `n == pairs.length`
• `1 <= n <= 1000`
• `-1000 <= lefti < righti <= 1000`

Code Examples

#1 Code Example with C Programming

```Code - C Programming```

``````
int cmp(const void *a, const void *b) {
int x = (*(int **)a)[0], y = (*(int **)b)[0];
//printf("cmp: %d:%d\n", x, y);

if (x < y) return -1;
else if (x > y) return 1;
return 0;
}
int findLongestChain(int** pairs, int pairsRowSize, int pairsColSize) {
int i, j, max, *dp;

dp = malloc(pairsRowSize * sizeof(int));
//assert(dp);

qsort(pairs, pairsRowSize, sizeof(int *), cmp);

max = 1;
for (i = 0; i < pairsRowSize; i ++) {
dp[i] = 1;
for (j = 0; j < i; j ++) {
if (pairs[i][0] > pairs[j][1] &&
dp[i] < dp[j] + 1) {
dp[i] = dp[j] + 1;
if (max < dp[i]) max = dp[i];
}
}
}

free(dp);

return max;
}
``````
Copy The Code &

Input

cmd
pairs = [[1,2],[2,3],[3,4]]

Output

cmd
2

#2 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int findLongestChain(int[][] pairs) {
Arrays.sort(pairs, new Comparator() {
@Override
public int compare(int[] o1, int[] o2) {
return o1[0] - o2[0];
}
});

int[] temp = new int[pairs.length];
Arrays.fill(temp, 1);
for (int i=1; i``````
``` Copy The Code & Input x – + cmd pairs = [[1,2],[2,3],[3,4]] Output x – + cmd 2 ```
``` #3 Code Example with Javascript Programming Code - Javascript Programming const findLongestChain = function(pairs) { pairs.sort((a, b) => a[1] - b[1]) let end = pairs[0][1], res = 1 for(let i = 1, len = pairs.length; i < len; i++) { if(pairs[i][0] > end) { res++ end = pairs[i][1] } } return res }; Copy The Code & Input x – + cmd pairs = [[1,2],[7,8],[4,5]] Output x – + cmd 3 #4 Code Example with Python Programming Code - Python Programming class Solution: def findLongestChain(self, pairs): pairs.sort(key = lambda x: x[1]) res, pre = 1, pairs[0][1] for c, d in pairs[1:]: if pre < c: pre = d res += 1 return res Copy The Code & Input x – + cmd pairs = [[1,2],[7,8],[4,5]] Output x – + cmd 3 Demonstration ```
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