{"id":6999,"date":"2020-06-28T01:25:06","date_gmt":"2020-06-28T08:25:06","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=6999"},"modified":"2020-06-28T01:35:40","modified_gmt":"2020-06-28T08:35:40","slug":"leetcode-1498-number-of-subsequences-that-satisfy-the-given-sum-condition","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/two-pointers\/leetcode-1498-number-of-subsequences-that-satisfy-the-given-sum-condition\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1498. Number of Subsequences That Satisfy the Given Sum Condition"},"content":{"rendered":"\n

Given an array of integers nums<\/code> and an integer target<\/code>.<\/p>\n\n\n\n

Return the number of non-empty<\/strong> subsequences of nums<\/code> such that the sum of the minimum and maximum element on it is less or equal than target<\/code>.<\/p>\n\n\n\n

Since the answer may be too large, return it modulo 10^9 + 7.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> nums = [3,5,6,7], target = 9\nOutput:<\/strong> 4\nExplanation: <\/strong>There are 4 subsequences that satisfy the condition.\n[3] -> Min value + max value <= target (3 + 3 <= 9)\n[3,5] -> (3 + 5 <= 9)\n[3,5,6] -> (3 + 6 <= 9)\n[3,6] -> (3 + 6 <= 9)\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [3,3,6,8], target = 10\nOutput:<\/strong> 6\nExplanation: <\/strong>There are 6 subsequences that satisfy the condition. (nums can have repeated numbers).\n[3] , [3] , [3,3], [3,6] , [3,6] , [3,3,6]<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [2,3,3,4,6,7], target = 12\nOutput:<\/strong> 61\nExplanation: <\/strong>There are 63 non-empty subsequences, two of them don't satisfy the condition ([6,7], [7]).\nNumber of valid subsequences (63 - 2 = 61).\n<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [5,2,4,1,7,6,8], target = 16\nOutput:<\/strong> 127\nExplanation: <\/strong>All non-empty subset satisfy the condition (2^7 - 1) = 127<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= nums.length <= 10^5<\/code><\/li>
  • 1 <= nums[i] <= 10^6<\/code><\/li>
  • 1 <= target <= 10^6<\/code><\/li><\/ul>\n\n\n\n
    Solution: Two Pointers<\/strong><\/h2>\n\n\n\n
    Since order of the elements in the subsequence doesn’t matter, we can sort the input array.
    Very similar to two sum, we use two pointers (i, j) to maintain a window, s.t. nums[i] +nums[j] <= target.
    Then fix nums[i], any subset of (nums[i+1~j]) gives us a valid subsequence, thus we have 2^(j-(i+1)+1) = 2^(j-i) valid subsequence for window (i, j).<\/p>\n\n\n\n
    Time complexity: O(nlogn) \/\/ Sort
    Space complexity: O(n) \/\/ need to precompute 2^n % kMod.<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int numSubseq(vector& nums, int target) {\n    constexpr int kMod = 1e9 + 7;\n    const int n = nums.size();\n    vector p(n + 1, 1);\n    for (int i = 1; i <= n; ++i) \n      p[i] = (p[i - 1] << 1) % kMod;\n    sort(begin(nums), end(nums));\n    int ans = 0;\n    for (int i = 0, j = n - 1; i <= j; ++i) {\n      while (i <= j && nums[i] + nums[j] > target) --j;\n      if (i > j) continue;\n      \/\/ In subarray nums[i~j]:\n      \/\/ min = nums[i], max = nums[j]\n      \/\/ nums[i] + nums[j] <= target\n      \/\/ {nums[i], (j - i - 1 + 1 values)}\n      \/\/ Any subset of the right part gives a valid subsequence \n      \/\/ in the original array. And There are 2^(j - i) ones.\n      ans = (ans + p[j - i]) % kMod;\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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