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[!Info] An array is a collection of the same type of data stored on a contiguous memory space

Binary Search

leetcode 704 Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

[!Tip] Preventing left + right is out of the range of int, so we have

int mid = left + ((right - left) / 2)

Left-closed and right-closed interval [left, right]

• left == right is meaning, so while(left <= right)
• Since right is valid index, if(midValue < target) right = mid - 1;

    while(left <= right)
    {
        int mid = left + ((right -left) >> 1);
        if(/*equal*/)
            return mid;
        else if(/*less*/)
            left = mid + 1;
        else // greater
            right = mid - 1;
    }
    return -1;
  

Left-closed, right-open Interval [left, right)

• left == right is invalid, so while(left < right)
• since right is invalid index, if(midValue < target) right = mid;

  		if(/*equal*/)
            return mid;
        else if(/*less*/)
            left = mid + 1;
        else // greater
            right = mid;
  

Extension Questions

35  Search Insert Position

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You must write an algorithm with O(log n) runtime complexity.

int searchInsert(vector<int>& nums, int target) {
	int left = 0, right = nums.size() - 1;
	while(left <= right)
	{
		int mid = left + ((right - left) >> 1);
		cout << left << right << endl;
		if(nums[mid] == target)
			return mid;
		else if(nums[mid] < target)
			left = mid + 1;
		else
			right = mid - 1;
	}

	// now l > r, three situations:
	// [0, -1] -> 0
	// [x, x - 1] -> x
	// [arraySize, arraySize - 1]
	return left;

}

34 Find First and Last Position of Elements in Sorted Array

Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

You must write an algorithm with O(log n) runtime complexity.

vector<int> searchRange(vector<int>& nums, int target) {
	int leftBorder = getBorder(nums, target, true);
	int rightBorder = getBorder(nums, target, false);
	if(leftBorder == -2 || rightBorder == -2)   return {-1, -1};
	if (rightBorder - leftBorder > 1) return {leftBorder + 1, rightBorder - 1};
	return {-1, -1};
}



/*
* -2 not set
*/

int getBorder(vector<int>& nums, int target, bool bLeftOrRight)
{
	int left = 0, right = nums.size() - 1;
	int border = -2; // not set
	while(left <= right)
	{
		int mid = left + ((right - left) >> 1);
		// cout << left << right << endl;
		if(!bLeftOrRight) // calc right border
		{
			if(nums[mid] > target)
				right = mid - 1;
			else // let left pointer move towards right border, so left move whenever nums[mid] <= target
			{
				left = mid + 1;
				border = left;
			}
		}

		else // calc for left border
		{
			if(nums[mid] < target)
				left = mid + 1;
			else // let right pointer move towards left border, so right move whenever nums[mid] >= target
			{
				right = mid - 1;
				border = right;
			}
		}
	}
	return border;
}

59 Spiral Matrix

Given a positive integer n, generate an n x n matrix filled with elements from 1 to n2 in spiral order.

[!Caution] Consider how to divide the intervals

1. 4 loops for four sides of a square
2. each loop is limited to a left-close right-open interval, the last number letting the next loop handle it.

![[Array-20240718174307980.webp]]

vector<vector<int>> generateMatrix(int n) {
	vector<vector<int>> res(n, vector<int>(n, 0)); // 2 dimension array
	int startx = 0, starty = 0; // start position of each loop
	int mid = n / 2; // center index
	int loop = n / 2; // number of loops, if n is odd, extra care at center
	int count = 1; // number to fill in
	int offset = 1; // limit the length of row and  columns
	int i,j;
	while (loop --) {
		i = startx;
		j = starty;
		auto setValueWithLog = [&res, &i, &j, &count]()
		{
			res[i][j] = count;
			cout << "Set(" << i << ", " << j << ") = " << count << endl;
			++count;
		};
		
		// Left-closed (draw), right-open(not draw, next for loop will fill it for us)
		// left -> right
		for (; j < n - offset; j++)
			setValueWithLog();
		cout << endl;
		
		// up -> down
		for (; i < n - offset; i++)
			setValueWithLog();
		cout << endl;

		// right -> left
		for (; j > starty; j--)
			setValueWithLog();
		cout << endl;

		// down -> up
		for (; i > startx; i--)
			setValueWithLog();

		// one loop finished, update the start of next loop
		++startx;
		++starty;
		++offset;
	}
	
	// if n is odd, extra care at the center
	if (n % 2) {
		res[mid][mid] = count;
	}
	return res;
}

Dual Pointer Method

[!Definition] A fast pointer and a slow pointer can do the work of two loops under one loop

Fast and Slow Pointers

27 Remove Element

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.

Consider the number of elements in nums which are not equal to val be k, to get accepted, you need to do the following things:

• Change the array nums such that the first k elements of nums contain the elements which are not equal to val. The remaining elements of nums are not important as well as the size of nums.
• Return k.

int removeElement(vector<int>& nums, int val) {
        int slow = 0;
        for (int fast = 0; fast < nums.size(); fast++) {
            if (val != nums[fast]) {
                nums[slow++] = nums[fast];
            }
        }
        return slow;
    }

Simulation:

f
3 2 2 3 `f++`
s

   f
3 2 2 3 `nums[f] != 3: nums[s++] = 2; f++`
s

    f
2 2 2  3 `nums[f] != 3; nums[s++] = 2; f++`
  s
   
      f
2 2 2 3 `f++`
    s

`return 2`

Front and Back Pointers

27 Remove Element

int removeElement(vector<int>& nums, int val) {
	int left = 0, right = nums.size() - 1;
	while(left <= right){
		if(nums[left] == val){
			nums[left] = nums[right];
			right--;
		}else {
			// 这里兼容了right指针指向的值与val相等的情况
			left++;
		}
	}
	return left;
}

Simulation:
      r
3 2 2 3  `nums[0] = nums[3] = 3`
l

    r
3 2 2 3 `nums[0] = nums[2] = 2`
l

   r
2 2 2 3
l

   r
2 2 2 3 `left == right: left++;`
   l
   
return ++1;

977 Square of sorted non-descending array

Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.

vector<int> sortedSquares(vector<int>& nums) {
	int l = 0, r = nums.size() - 1;
	vector<int> res(r + 1);
	int arrayIndex = r;
	while(l <= r)
	{
		if(abs(nums[l]) >= abs(nums[r]))
		{
			res[arrayIndex--] = nums[l] * nums[l];
			++l;
		}
		else
		{
			res[arrayIndex--] = nums[r] * nums[r];
			--r;
		}
	}
	return res;
}

Simulation:
1.
          r
-4 -1 0 3 10  `(-4)^2 < 10^2 : res[i--] = (10^2); r--`
l

2.             res:
        r    |       i     |
-4 -1 0 3 10 | _ _ _ _ 100 |`(-4)^2 > 3^2 : res[i--]=(-4)^2; l++`
l            |             |

3.             res:
        r    |     i        |
-4 -1 0 3 10 | _ _ _ 16 100 |`(-1)^2 < 3^2 : res[i--]=(3)^2; r--`
    l        |              |

4.             res
      r      |   i          |
-4 -1 0 3 10 | _ _ 9 16 100 |`(-1)^2 > 0^2 : res[i--]=(-1)^2; l++`
    l        |              |

5.             res
      r      | i            |
-4 -1 0 3 10 | _ 1 9 16 100 |`(0)^2 > 0^2 : res[i--]=(0)^2; l++`
      l      |              |

res:
0 1 9 16 100

Sliding Window

[!important] The so-called sliding window is a constant adjustment of the starting and ending positions of the subsequence to arrive at the result we want.

209 Minimum Size Subarray Sum

Given an array of positive integers nums and a positive integer target, return the minimal length of a subarray whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.  

int minSubArrayLen(int target, vector<int>& nums) {
	int l = 0, r = 0;
	int minLen = INT32_MAX;
	int sum = 0;
	while(r < nums.size())
	{
		sum += nums[r];
		while(l <= r && sum >= target)
		{
			if(minLen > (r - l) + 1)
				minLen = r - l + 1;
			sum -= nums[l++];
		}
		++r;
	}
	return minLen == INT32_MAX ? 0 : minLen;
}

Simulation:
target = 7

r
2 3 1 2 4 3
l

      r
2 3 1 2 4 3 `sum = 8 > target: minlen = 4;`
l