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.length1 <= 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;
}
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int findLongestChain(int[][] pairs) {
Arrays.sort(pairs, new Comparator<int[]>() {
@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 < pairs.length; i++) {
for (int j=0; j < i; j++) {
if (pairs[j][1] < pairs[i][0]) {
temp[i] = Math.max(temp[i], temp[j] + 1);
}
}
}
return Arrays.stream(temp).max().getAsInt();
}
}
Input
Output
#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
};
Input
Output
#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
Input
Output