Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1906.html

1906. Minimum Absolute Difference Queries

Level

Medium

Description

The minimum absolute difference of an array a is defined as the minimum value of |a[i] - a[j]|, where 0 <= i < j < a.length and a[i] != a[j]. If all elements of a are the same, the minimum absolute difference is -1.

  • For example, the minimum absolute difference of the array [5,2,3,7,2] is |2 - 3| = 1. Note that it is not 0 because a[i] and a[j] must be different.

You are given an integer array nums and the array queries where queries[i] = [l_i, r_i]. For each query i, compute the minimum absolute difference of the subarray nums[l_i...r_i] containing the elements of nums between the 0-based indices l_i and r_i (inclusive).

Return an array ans where ans[i] is the answer to the i-th query.

A subarray is a contiguous sequence of elements in an array.

The value of |x| is defined as:

  • x if x >= 0.
  • -x if x < 0.

Example 1:

Input: nums = [1,3,4,8], queries = [[0,1],[1,2],[2,3],[0,3]]

Output: [2,1,4,1]

Explanation: The queries are processed as follows:

  • queries[0] = [0,1]: The subarray is [1,3] and the minimum absolute difference is 1-3 = 2.
  • queries[1] = [1,2]: The subarray is [3,4] and the minimum absolute difference is 3-4 = 1.
  • queries[2] = [2,3]: The subarray is [4,8] and the minimum absolute difference is 4-8 = 4.
  • queries[3] = [0,3]: The subarray is [1,3,4,8] and the minimum absolute difference is 3-4 = 1.

Example 2:

Input: nums = [4,5,2,2,7,10], queries = [[2,3],[0,2],[0,5],[3,5]]

Output: [-1,1,1,3]

Explanation: The queries are processed as follows:

  • queries[0] = [2,3]: The subarray is [2,2] and the minimum absolute difference is -1 because all the elements are the same.
  • queries[1] = [0,2]: The subarray is [4,5,2] and the minimum absolute difference is 4-5 = 1.
  • queries[2] = [0,5]: The subarray is [4,5,2,2,7,10] and the minimum absolute difference is 4-5 = 1.
  • queries[3] = [3,5]: The subarray is [2,7,10] and the minimum absolute difference is 7-10 = 3.

Constraints:

  • 2 <= nums.length <= 10^5
  • 1 <= nums[i] <= 100
  • 1 <= queries.length <= 2 * 10^4
  • 0 <= l_i < r_i < nums.length

Solution

Since the number range of nums is limited, store the number of occurrences for each number from 1 to 100 for each prefix of nums. Then for each query, the number of occurrences for each number from 1 to 100 in the query range can be calculated efficiently, and the minimum absolute difference can be calculated by looping over the numbers from 1 to 100 and calculating the differences between each paif of adjacent numbers that both occur in the range.

  • class Solution {
    public int[] minDifference(int[] nums, int[][] queries) {
        Map<Integer, int[]> map = new HashMap<Integer, int[]>();
        int length = nums.length;
        int[] counts = new int[101];
        for (int i = 0; i < length; i++) {
            int num = nums[i];
            counts[num]++;
            int[] copy = new int[101];
            System.arraycopy(counts, 0, copy, 0, 101);
            map.put(i, copy);
        }
        int queriesCount = queries.length;
        int[] ans = new int[queriesCount];
        for (int i = 0; i < queriesCount; i++) {
            int[] query = queries[i];
            int[] rangeCounts = getRangeCounts(map, query[0], query[1]);
            int curMin = Integer.MAX_VALUE;
            int prev = -1;
            for (int j = 1; j <= 100; j++) {
                if (rangeCounts[j] > 0) {
                    if (prev > 0) {
                        int diff = j - prev;
                        curMin = Math.min(curMin, diff);
                    }
                    prev = j;
                }
            }
            ans[i] = curMin == Integer.MAX_VALUE ? -1 : curMin;
        }
        return ans;
    }

    public int[] getRangeCounts(Map<Integer, int[]> map, int start, int end) {
        int[] endCounts = map.get(end);
        if (start == 0)
            return endCounts;
        int[] prevCounts = map.get(start - 1);
        int[] rangeCounts = new int[101];
        for (int i = 0; i <= 100; i++)
            rangeCounts[i] = endCounts[i] - prevCounts[i];
        return rangeCounts;
    }
}

############

class Solution {
    public int[] minDifference(int[] nums, int[][] queries) {
        int m = nums.length, n = queries.length;
        int[][] preSum = new int[m + 1][101];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= 100; ++j) {
                int t = nums[i - 1] == j ? 1 : 0;
                preSum[i][j] = preSum[i - 1][j] + t;
            }
        }

        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int left = queries[i][0], right = queries[i][1] + 1;
            int t = Integer.MAX_VALUE;
            int last = -1;
            for (int j = 1; j <= 100; ++j) {
                if (preSum[right][j] > preSum[left][j]) {
                    if (last != -1) {
                        t = Math.min(t, j - last);
                    }
                    last = j;
                }
            }
            if (t == Integer.MAX_VALUE) {
                t = -1;
            }
            ans[i] = t;
        }
        return ans;
    }
}

  • class Solution:
    def minDifference(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        m, n = len(nums), len(queries)
        pre_sum = [[0] * 101 for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, 101):
                t = 1 if nums[i - 1] == j else 0
                pre_sum[i][j] = pre_sum[i - 1][j] + t

        ans = []
        for i in range(n):
            left, right = queries[i][0], queries[i][1] + 1
            t = inf
            last = -1
            for j in range(1, 101):
                if pre_sum[right][j] - pre_sum[left][j] > 0:
                    if last != -1:
                        t = min(t, j - last)
                    last = j
            if t == inf:
                t = -1
            ans.append(t)
        return ans

############

# 1906. Minimum Absolute Difference Queries
# https://leetcode.com/problems/minimum-absolute-difference-queries

class Solution:
    def minDifference(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        dp = [[0] * (max(nums) + 1)]
        
        for i, x in enumerate(nums):
            t = dp[-1][:]
            t[x] += 1
            dp.append(t)
        
        res = []
        for start, end in queries:
            A = [i for i, (a,b) in enumerate(zip(dp[start], dp[end + 1])) if a != b]
            res.append(min([b - a for a,b in zip(A, A[1:])] or [-1]))
            
        return res


  • class Solution {
public:
    vector<int> minDifference(vector<int>& nums, vector<vector<int>>& queries) {
        int m = nums.size(), n = queries.size();
        int preSum[m + 1][101];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= 100; ++j) {
                int t = nums[i - 1] == j ? 1 : 0;
                preSum[i][j] = preSum[i - 1][j] + t;
            }
        }

        vector<int> ans(n);
        for (int i = 0; i < n; ++i) {
            int left = queries[i][0], right = queries[i][1] + 1;
            int t = 101;
            int last = -1;
            for (int j = 1; j <= 100; ++j) {
                if (preSum[right][j] > preSum[left][j]) {
                    if (last != -1) {
                        t = min(t, j - last);
                    }
                    last = j;
                }
            }
            if (t == 101) {
                t = -1;
            }
            ans[i] = t;
        }
        return ans;
    }
};

  • func minDifference(nums []int, queries [][]int) []int {
	m, n := len(nums), len(queries)
	preSum := make([][101]int, m+1)
	for i := 1; i <= m; i++ {
		for j := 1; j <= 100; j++ {
			t := 0
			if nums[i-1] == j {
				t = 1
			}
			preSum[i][j] = preSum[i-1][j] + t
		}
	}

	ans := make([]int, n)
	for i := 0; i < n; i++ {
		left, right := queries[i][0], queries[i][1]+1
		t, last := 101, -1
		for j := 1; j <= 100; j++ {
			if preSum[right][j]-preSum[left][j] > 0 {
				if last != -1 {
					if t > j-last {
						t = j - last
					}
				}
				last = j
			}
		}
		if t == 101 {
			t = -1
		}
		ans[i] = t
	}
	return ans
}

  • function minDifference(nums: number[], queries: number[][]): number[] {
    let m = nums.length,
        n = queries.length;
    let max = 100;
    // let max = Math.max(...nums);
    let pre: number[][] = [];
    pre.push(new Array(max + 1).fill(0));
    for (let i = 0; i < m; ++i) {
        let num = nums[i];
        pre.push(pre[i].slice());
        pre[i + 1][num] += 1;
    }

    let ans = [];
    for (let [left, right] of queries) {
        let last = -1;
        let min = Infinity;
        for (let j = 1; j < max + 1; ++j) {
            if (pre[left][j] < pre[right + 1][j]) {
                if (last != -1) {
                    min = Math.min(min, j - last);
                }
                last = j;
            }
        }
        ans.push(min == Infinity ? -1 : min);
    }
    return ans;
}

All Problems

All Solutions