Minimum Total Distance Traveled

Solution For Minimum Total Distance Traveled

Problem Statement:

Given an array of n integer coordinates where A[i] represents the coordinate of the ith point. You are allowed to move any number of times, but only in a straight line. What is the minimum total distance you have to travel to reach all points?

Input: [[1,1],[3,4],[-1,0]] Output: 7

One optimal solution: 1 -> 3 -> -1 -> 1: distance = 2 + 5 + 4 = 11


The given problem can be solved using the greedy algorithm. We need to find the next point that is closest to the current point and go to that point. Repeat this process till we cover all points.

Let the current position be represented by the variable curr and let the total distance traveled so far be represented by the variable dist. We can keep track of the current position and the total distance using these variables.

For each point in the given array, we need to find the Euclidean distance between the current point and all other remaining points. We can use the Euclidean distance formula to calculate the distance between two points, which is sqrt(pow(x2-x1, 2) + pow(y2-y1, 2)).

We can store the distances in a priority queue in ascending order. We can then select the point with the smallest distance from the priority queue, travel to that point, and update the current position and total distance. We then repeat this process till all points are visited.


  1. Initialize dist and curr variables as 0 and the first point in the array.
  2. Create an empty priority queue and insert all the points of the array, except the first one.
  3. While the priority queue is not empty, do the following:
    a. Remove the point with the smallest distance from the priority queue.
    b. Calculate the Euclidean distance between the current point curr and the removed point.
    c. Add the distance to dist.
    d. Update curr to the removed point.
    e. Remove the removed point from the priority queue.
  4. Return the total distance dist.

Time Complexity:

The time complexity of this algorithm is O(n^2 log n), where n is the number of points in the array. The reason is that, for each point, we are computing the Euclidean distance to all other remaining points, which is an O(n) operation. We are then adding and removing elements from the priority queue, which takes O(log n) time. Therefore, the overall time complexity is O(n^2 log n).


Here is the complete solution in Python:

import math
import queue

def minTotalDistance(points: List[List[int]]) -> int:
n = len(points)
dist = 0
curr = points[0] pq = queue.PriorityQueue()

# insert all points except the first one
for i in range(1, n):
    pq.put((math.dist(curr, points[i]), points[i]))

# remove points from the priority queue and calculate the total distance
while not pq.empty():
    p = pq.get()[1]
    dist += math.dist(curr, p)
    curr = p

return int(dist)

Note: Since the result is a floating-point number, we convert it to an integer before returning. The math.dist function is used to calculate the Euclidean distance between two points.

Step by Step Implementation For Minimum Total Distance Traveled

 * Given an array of points on a grid, find the minimum total distance that must be traveled in order to visit all points.
 * The solution is to find the shortest path from the starting point to the end point that visits all points in between.
 * To do this, we can use a breadth-first search algorithm. We start at the starting point, and add all adjacent points
 * to our queue. Then, we visit the next point in our queue, and add all of its adjacent points to the queue. We continue
 * this process until we reach the end point.
 * The shortest path will be the path that visits the fewest points. Therefore, we can keep track of the number of points
 * visited in each path, and return the path with the smallest number of points.

public int minTotalDistance(int[][] points) {
    if (points == null || points.length == 0) {
        return 0;
    // we use a queue to store the points we need to visit
    Queue queue = new LinkedList<>();
    // we use a set to keep track of the points we've already visited
    Set visited = new HashSet<>();
    // we start at the first point
    int[] start = points[0];
    // we keep track of the shortest path
    int shortestPath = Integer.MAX_VALUE;
    while (!queue.isEmpty()) {
        int[] point = queue.poll();
        int x = point[0];
        int y = point[1];
        // we've reached the end point, so we update the shortest path
        if (x == points[points.length - 1][0] && y == points[points.length - 1][1]) {
            shortestPath = Math.min(shortestPath, visited.size());
        // add all adjacent points to the queue
        addAdjacentPoints(queue, visited, x + 1, y);
        addAdjacentPoints(queue, visited, x - 1, y);
        addAdjacentPoints(queue, visited, x, y + 1);
        addAdjacentPoints(queue, visited, x, y - 1);
    return shortestPath;

// helper function to add adjacent points to the queue
private void addAdjacentPoints(Queue queue, Set visited, int x, int y) {
    // we only add points that are within the grid and that we haven't visited yet
    if (x >= 0 && x < points.length && y >= 0 && y < points[0].length && !visited.contains(getPointString(x, y))) {
        queue.add(new int[] {x, y});
        visited.add(getPointString(x, y));

// helper function to get a string representation of a point
private String getPointString(int[] point) {
    return getPointString(point[0], point[1]);

private String getPointString(int x, int y) {
    return x + "," + y;
This problem can be solved using a dynamic programming approach.

We can create a 2D array, where each element represents the minimum total distance that can be traveled to reach that point.

We can then fill in this array using the following recursive relation:

min_distance[i][j] = min(min_distance[i-1][j], min_distance[i][j-1]) + grid[i][j]

where grid is the given 2D array of distances.

The base cases are as follows:

min_distance[0][0] = grid[0][0]

min_distance[i][0] = min_distance[i-1][0] + grid[i][0]

min_distance[0][j] = min_distance[0][j-1] + grid[0][j]

The final answer is the value in min_distance[n-1][m-1], where n and m are the dimensions of the grid.
var minimumTotalDistance = function(arr) {
    let total = 0;
    let min = Infinity;
    let max = -Infinity;
    for (let i=0; i max) {
            max = arr[i];
    let avg = Math.floor(total / arr.length);
    let left = 0;
    let right = arr.length - 1;
    while (left < right) {
        while (arr[left] < avg) {
        while (arr[right] > avg) {
        if (left < right) {
            let temp = arr[left];
            arr[left] = arr[right];
            arr[right] = temp;
    let mid = Math.floor(arr.length / 2);
    let median = arr[mid];
    if (arr.length % 2 === 0) {
        median = (arr[mid] + arr[mid-1]) / 2;
    let leftSum = 0;
    let rightSum = 0;
    for (let i=0; i
int minimumTotal(vector>& triangle) {
        int n = triangle.size();
        vector minlen(triangle.back());
        for (int layer = n-2; layer >= 0; layer--) // For each layer
            for (int i = 0; i <= layer; i++) // Check its every 'node'
                // Find the lesser of its two children, and sum the current value in the triangle with it.
                minlen[i] = min(minlen[i], minlen[i+1]) + triangle[layer][i]; 
        return minlen[0];
Assuming we have a 2D array of points, we can find the minimum total distance traveled by iterating through each point and calculating the distance to all other points. To find the distance between two points, we can use the Euclidean distance formula.

public int MinTotalDistance(int[][] points) { int minDistance = int.MaxValue; // Iterate through all points for (int i = 0; i < points.Length; i++) { int currDistance = 0; // Iterate through all other points and calculate distance for (int j = 0; j < points.Length; j++) { if (i != j) { currDistance += Distance(points[i], points[j]); } } // Update minimum distance if necessary minDistance = Math.Min(minDistance, currDistance); } return minDistance; } private int Distance(int[] point1, int[] point2) { return (int)Math.Sqrt(Math.Pow(point1[0] - point2[0], 2) + Math.Pow(point1[1] - point2[1], 2)); }

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